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Atrium > Mathematics, Physics & Philosophy: improper integrals

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- CommentRowNumber1.
- CommentAuthorMike Shulman
- CommentTimeMar 1st 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexHere we go again. My calculus book defines the Riemann integral for arbitrary functions (not necessarily continuous), and states a theorem that continuous functions are integrable. Then later on it defines improper integrals, e.g. $$\int_a^b f(x) dx = \lim_{t\to b^-} \int_a^t f(x) dx$$ if $f$ is continuous on $[a,b)$ but not at $b$. This is logically unsound because we're giving the symbol $\int_a^b f(x) dx$ two different meanings for a not-necessarily-continuous $f$, but it doesn't seem to cause problems in practice. For one thing, if $f$ *is* continuous on $[a,b]$, then the above equality is a true statement (use FTC and the fact that any antiderivative of $f$, being differentiable, is continuous). For another, we usually only apply improper integrals to unbounded functions, which are not Riemann integrable in the usual sense. But I wonder whether it is *always* consistent? In other words, suppose $f$ is Riemann integrable on $[a,t]$ for all $t\in [a,b]$. Is it necessarily true that $\int_a^b f(x) dx = \lim_{t\to b^-} \int_a^t f(x) dx$?

Here we go again.

My calculus book defines the Riemann integral for arbitrary functions (not necessarily continuous), and states a theorem that continuous functions are integrable. Then later on it defines improper integrals, e.g.
$\int_a^b f(x) dx = \lim_{t\to b^-} \int_a^t f(x) dx$
if $f$ is continuous on $[a,b)$ but not at $b$ . This is logically unsound because we’re giving the symbol $\int_a^b f(x) dx$ two different meanings for a not-necessarily-continuous $f$ , but it doesn’t seem to cause problems in practice. For one thing, if $f$ is continuous on $[a,b]$ , then the above equality is a true statement (use FTC and the fact that any antiderivative of $f$ , being differentiable, is continuous). For another, we usually only apply improper integrals to unbounded functions, which are not Riemann integrable in the usual sense. But I wonder whether it is always consistent?

In other words, suppose $f$ is Riemann integrable on $[a,t]$ for all $t\in [a,b]$ . Is it necessarily true that $\int_a^b f(x) dx = \lim_{t\to b^-} \int_a^t f(x) dx$ ?
- CommentRowNumber2.
- CommentAuthorTobyBartels
- CommentTimeMar 1st 2014
- PermaLink
Author: TobyBartels
Format: MarkdownItexH\'m, my textbook does the same thing. That does seem odd. In class, I defined improper integrals only when $f$ is actually *undefined* at some point in $[a,b]$ (which is automatically true if $a = -\infty$ or $b = \infty$), but I didn\'t require continuity elsewhere. This covers every improper integral that appears in any examples or exercises (so the book never considers a function that is discontinuous somewhere that it is defined, at least not when discussing improper integrals). Also, all of the theorems stated in the book for improper integrals (the Direct and Limit Comparison Tests, and the Integral Test for infinite series) are true of it (which I know because those theorems only deal with nonnegative functions and therefore I can pass to the Lebesgue integral, where they are easy). I should tell the students that my definition is not quite equivalent to the book's, even though they agree when both apply. (I *did* notice that the Integral Test required a continuous function, and I did point out to them that this clause is unnecessary; this appears to be the same issue.) We are saved by the theory of Henstock integrals. These agree with Riemann integrals whenever the latter are defined, and they obey Hake\'s Theorem, which states precisely that the equation that you want holds whenever *either* side is defined. Thus even for Riemann integrals, the equation is true whenever *both* sides are defined. (I don\'t know why people don\'t call that Hake\'s Theorem too, but I can't get any hits on Google for a theorem with that name that aren\'t about the Henstock integral under one of its many names.)

H'm, my textbook does the same thing. That does seem odd.

In class, I defined improper integrals only when $f$ is actually undefined at some point in $[a,b]$ (which is automatically true if $a = -\infty$ or $b = \infty$ ), but I didn't require continuity elsewhere. This covers every improper integral that appears in any examples or exercises (so the book never considers a function that is discontinuous somewhere that it is defined, at least not when discussing improper integrals). Also, all of the theorems stated in the book for improper integrals (the Direct and Limit Comparison Tests, and the Integral Test for infinite series) are true of it (which I know because those theorems only deal with nonnegative functions and therefore I can pass to the Lebesgue integral, where they are easy). I should tell the students that my definition is not quite equivalent to the book’s, even though they agree when both apply. (I did notice that the Integral Test required a continuous function, and I did point out to them that this clause is unnecessary; this appears to be the same issue.)

We are saved by the theory of Henstock integrals. These agree with Riemann integrals whenever the latter are defined, and they obey Hake's Theorem, which states precisely that the equation that you want holds whenever either side is defined. Thus even for Riemann integrals, the equation is true whenever both sides are defined. (I don't know why people don't call that Hake's Theorem too, but I can’t get any hits on Google for a theorem with that name that aren't about the Henstock integral under one of its many names.)
- CommentRowNumber3.
- CommentAuthorMike Shulman
- CommentTimeMar 1st 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexThanks for pointing out Hake's Theorem! I had seen it referred to for Henstock integrals, but didn't make the connection that it would say something about Riemann integrals. It seems to me that we can say even more about the Riemann integral: if, as I assumed, $f$ is Riemann integrable on $[a,t]$ for all $t\in [a,b]$, then all those Riemann integrals agree with the Henstock integrals, and so by Hake's theorem for Henstock integrals, the limit on the RHS of my equation *does* exist (and is equal to the LHS). Moreover, isn't it true that a Riemann integrable function on a closed interval is also integrable on every subinterval, so that the equation holds for Riemann integrals whenever the LHS is defined?

Thanks for pointing out Hake’s Theorem! I had seen it referred to for Henstock integrals, but didn’t make the connection that it would say something about Riemann integrals. It seems to me that we can say even more about the Riemann integral: if, as I assumed, $f$ is Riemann integrable on $[a,t]$ for all $t\in [a,b]$ , then all those Riemann integrals agree with the Henstock integrals, and so by Hake’s theorem for Henstock integrals, the limit on the RHS of my equation does exist (and is equal to the LHS). Moreover, isn’t it true that a Riemann integrable function on a closed interval is also integrable on every subinterval, so that the equation holds for Riemann integrals whenever the LHS is defined?
- CommentRowNumber4.
- CommentAuthorMike Shulman
- CommentTimeMar 1st 2014
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Author: Mike Shulman
Format: MarkdownItexIt's a shame that the Henstock integral doesn't work directly over infinite intervals.

It’s a shame that the Henstock integral doesn’t work directly over infinite intervals.
- CommentRowNumber5.
- CommentAuthorTodd_Trimble
- CommentTimeMar 1st 2014
- (edited Mar 1st 2014)
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Author: Todd_Trimble
Format: MarkdownItex> Moreover, isn’t it true that a Riemann integrable function on a closed interval is also integrable on every subinterval, so that the equation holds for Riemann integrals whenever the LHS is defined? Yes. One thing that's sometimes helpful is that the class of Riemann-integrable functions on $[a, b]$ is the class of bounded functions that are continuous at all but a set of points of measure zero, and there the Riemann integral $\int_a^b f(x)\; d x$ agrees with the Lebesgue integral. (I forgot to say 'bounded' when I first wrote this, so I hope I remembered right. Can't find my source, Measure and Integral by Wheeden and Zygmund, to double-check.)

Moreover, isn’t it true that a Riemann integrable function on a closed interval is also integrable on every subinterval, so that the equation holds for Riemann integrals whenever the LHS is defined?

Yes. One thing that’s sometimes helpful is that the class of Riemann-integrable functions on $[a, b]$ is the class of bounded functions that are continuous at all but a set of points of measure zero, and there the Riemann integral $\int_a^b f(x)\; d x$ agrees with the Lebesgue integral. (I forgot to say ’bounded’ when I first wrote this, so I hope I remembered right. Can’t find my source, Measure and Integral by Wheeden and Zygmund, to double-check.)
- CommentRowNumber6.
- CommentAuthorMike Shulman
- CommentTimeMar 1st 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItex[Wikipedia](http://en.wikipedia.org/wiki/Riemann_integral#Integrability) agrees.

Wikipedia agrees.
- CommentRowNumber7.
- CommentAuthorMike Shulman
- CommentTimeMar 2nd 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexThis discussion makes me strongly tempted to "use" Henstock integrals in teaching calculus (for some meaning of "use"). The actual definition is too complicated to discuss, of course; but in my calculus classes, at least, we don't really use the definition of the Riemann integral at all anyway. In particular, we don't really talk about the meaning of taking the "limit as the mesh goes to 0" (since we haven't really talked about epsilons and deltas for ordinary limits), so I wouldn't really mind pretending in my head that the Henstock definition is just the way that we make that precise. And it would make me much happier to state Hake's theorem as a *theorem* which enables us to calculate some integrals that the FTC doesn't apply to directly, rather than having to introduce a new *definition* of the meaning of the integral of a previously non-integrable function. (The relaxed hypotheses on the FTC itself are an added bonus.)

This discussion makes me strongly tempted to “use” Henstock integrals in teaching calculus (for some meaning of “use”). The actual definition is too complicated to discuss, of course; but in my calculus classes, at least, we don’t really use the definition of the Riemann integral at all anyway. In particular, we don’t really talk about the meaning of taking the “limit as the mesh goes to 0” (since we haven’t really talked about epsilons and deltas for ordinary limits), so I wouldn’t really mind pretending in my head that the Henstock definition is just the way that we make that precise. And it would make me much happier to state Hake’s theorem as a theorem which enables us to calculate some integrals that the FTC doesn’t apply to directly, rather than having to introduce a new definition of the meaning of the integral of a previously non-integrable function. (The relaxed hypotheses on the FTC itself are an added bonus.)
- CommentRowNumber8.
- CommentAuthorTodd_Trimble
- CommentTimeMar 2nd 2014
- PermaLink
Author: Todd_Trimble
Format: MarkdownItexMike, there's an article in the American Mathematical Monthly which may be of interest to you: * Russell A. Gordon, _The Use of Tagged Partitions in Elementary Real Analysis_, AMM Vol. 105 No. 2 (February 1998), 107-117.
Mike, there’s an article in the American Mathematical Monthly which may be of interest to you:
- Russell A. Gordon, The Use of Tagged Partitions in Elementary Real Analysis, AMM Vol. 105 No. 2 (February 1998), 107-117.
- CommentRowNumber9.
- CommentAuthorMike Shulman
- CommentTimeMar 2nd 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexThanks, I'll have a look when I get a chance.

Thanks, I’ll have a look when I get a chance.
- CommentRowNumber10.
- CommentAuthorDavidRoberts
- CommentTimeMar 3rd 2014
- PermaLink
Author: DavidRoberts
Format: MarkdownItex[The Use of Tagged Partitions in Elementary Real Analysis](http://www.jstor.org/stable/2589642), for posterity.

The Use of Tagged Partitions in Elementary Real Analysis, for posterity.
- CommentRowNumber11.
- CommentAuthorDavidRoberts
- CommentTimeMar 3rd 2014
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Author: DavidRoberts
Format: MarkdownItexAnd that's a really nice article, by the way!

And that’s a really nice article, by the way!
- CommentRowNumber12.
- CommentAuthorMike Shulman
- CommentTimeMar 3rd 2014
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Author: Mike Shulman
Format: MarkdownItexIt is a nice article! It also made me realize how badly I've been infected with constructivism, as I found it painful to have to read such blatant proofs by contradiction. (-: I wonder whether there is any constructive version of the theory of tagged partitions?

It is a nice article! It also made me realize how badly I’ve been infected with constructivism, as I found it painful to have to read such blatant proofs by contradiction. (-: I wonder whether there is any constructive version of the theory of tagged partitions?
- CommentRowNumber13.
- CommentAuthorTodd_Trimble
- CommentTimeMar 4th 2014
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Author: Todd_Trimble
Format: MarkdownItexIt might be worthwhile going through it and rewriting, because most mathematicians are infected by lazy habits of blatantly overusing "proofs by contradiction" when direct proofs are available. Of course, that's easy for me to say, since I had no plans to do such a thing. :-)

It might be worthwhile going through it and rewriting, because most mathematicians are infected by lazy habits of blatantly overusing “proofs by contradiction” when direct proofs are available. Of course, that’s easy for me to say, since I had no plans to do such a thing. :-)
- CommentRowNumber14.
- CommentAuthorMike Shulman
- CommentTimeMar 4th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexA lot of the other theorems in that paper require modification to become true constructively. In particular, constructively the interval $[0,1]$ may not be covering-compact as a topological space, so it seems that $\delta$-fine tagged partitions may not exist. On the other hand, $[0,1]$ is compact as a locale, but it seems less likely that that sort of "point-free" compactness can be rephrased using tagged partitions, since they are very "point-ful".

A lot of the other theorems in that paper require modification to become true constructively. In particular, constructively the interval $[0,1]$ may not be covering-compact as a topological space, so it seems that $\delta$ -fine tagged partitions may not exist. On the other hand, $[0,1]$ is compact as a locale, but it seems less likely that that sort of “point-free” compactness can be rephrased using tagged partitions, since they are very “point-ful”.
- CommentRowNumber15.
- CommentAuthorspitters
- CommentTimeMar 4th 2014
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Author: spitters
Format: MarkdownItexI worked a bit on a construstive version of these integrals some 15 years ago. Indeed, I did not see how to do it in a pointwise way. I never tried to do it in a pointfree way. I'll try to find my notes later.

I worked a bit on a construstive version of these integrals some 15 years ago. Indeed, I did not see how to do it in a pointwise way. I never tried to do it in a pointfree way. I’ll try to find my notes later.
- CommentRowNumber16.
- CommentAuthorspitters
- CommentTimeMar 4th 2014
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Author: spitters
Format: MarkdownItexThe first problem one finds is that if the function $\delta$ (for fineness) is total (and definable), then it is uniformly continuous and its infimum is >0. We have retrieve the Riemann integral! So $\delta$ needs to be partial, but defined on a large set. Now if we consider $\delta_1$ and $\delta_2$ fine meshes, we want to take their infimum, but is it sufficient to only consider the infimum on the intersection of their domains ... I tried a number of possibilities in the past, but it does not seem to become very nice. I prefer algebraic integration theory where we mod out by the Null sets from the start.

The first problem one finds is that if the function $\delta$ (for fineness) is total (and definable), then it is uniformly continuous and its infimum is >0. We have retrieve the Riemann integral!

So $\delta$ needs to be partial, but defined on a large set. Now if we consider $\delta_1$ and $\delta_2$ fine meshes, we want to take their infimum, but is it sufficient to only consider the infimum on the intersection of their domains …

I tried a number of possibilities in the past, but it does not seem to become very nice.

I prefer algebraic integration theory where we mod out by the Null sets from the start.
- CommentRowNumber17.
- CommentAuthorMike Shulman
- CommentTimeMar 4th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexAnother reason that going pointfree would be problematic is that it's hard to define discontinuous functions in that setting. (-:

Another reason that going pointfree would be problematic is that it’s hard to define discontinuous functions in that setting. (-:
- CommentRowNumber18.
- CommentAuthorspitters
- CommentTimeMar 5th 2014
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Author: spitters
Format: MarkdownItexOne further thought. I believe it was Halmos who emphasized that the goal of integration is not to make as many functions integrable, but to have an integral with good closure properties. IIRC, for the Henstock integral, we do not have: $f$ integrable implies $|f|$ integrable. This makes it less suitable for functional analysis.

One further thought. I believe it was Halmos who emphasized that the goal of integration is not to make as many functions integrable, but to have an integral with good closure properties. IIRC, for the Henstock integral, we do not have: $f$ integrable implies $|f|$ integrable. This makes it less suitable for functional analysis.
- CommentRowNumber19.
- CommentAuthorMike Shulman
- CommentTimeMar 6th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexWe do, however, have a better FTC, plus Hake's theorem, as discussed in the [other thread](http://nforum.mathforge.org/discussion/5728/henstock-integral). I think those are more important advantages of it than the fact that it integrates "more" functions than the Lebesgue integral does. So it depends on what properties you consider more important. On further thought, I don't quite understand the point in #16. It's not a theorem that every total function is uniformly continuous. It's constructively consistent to *assume* that every total function is uniformly continuous, but in that case, every function would be Riemann integrable anyway (right?), so there would be no problem. The Henstock integral is only interesting if we don't assume every function is continuous. It seems to me that a good way to start would be to look at the gauges used in the proofs of the properties of the Henstock integral and see what they look like constructively. For instance, in the paper mentioned above, there is the following example considered as a beginning of a proof of FTC: for a differentiable function $f$ and $\epsilon\gt 0$, there is a gauge such that if ${|x-a|}\lt \delta(a)$, then ${|f(x)-f(a)-f'(a)(x-a)|} \lt \epsilon {|x-a|}$. Now differentiability gives us existence of some such $\delta(a)$, but to get a function $\delta$, we seem to need the axiom of choice. However, without AC, it appears that $\delta$ can be regarded as a function into the [[lower real number|lower reals]]. So one natural thing to try would be to define gauges to be functions into the lower reals. But I haven't looked at the proof of the other properties of the Henstock integral to see what kind of gauges they use.

We do, however, have a better FTC, plus Hake’s theorem, as discussed in the other thread. I think those are more important advantages of it than the fact that it integrates “more” functions than the Lebesgue integral does. So it depends on what properties you consider more important.

On further thought, I don’t quite understand the point in #16. It’s not a theorem that every total function is uniformly continuous. It’s constructively consistent to assume that every total function is uniformly continuous, but in that case, every function would be Riemann integrable anyway (right?), so there would be no problem. The Henstock integral is only interesting if we don’t assume every function is continuous.

It seems to me that a good way to start would be to look at the gauges used in the proofs of the properties of the Henstock integral and see what they look like constructively. For instance, in the paper mentioned above, there is the following example considered as a beginning of a proof of FTC: for a differentiable function $f$ and $\epsilon\gt 0$ , there is a gauge such that if ${|x-a|}\lt \delta(a)$ , then ${|f(x)-f(a)-f'(a)(x-a)|} \lt \epsilon {|x-a|}$ . Now differentiability gives us existence of some such $\delta(a)$ , but to get a function $\delta$ , we seem to need the axiom of choice. However, without AC, it appears that $\delta$ can be regarded as a function into the lower reals. So one natural thing to try would be to define gauges to be functions into the lower reals. But I haven’t looked at the proof of the other properties of the Henstock integral to see what kind of gauges they use.
- CommentRowNumber20.
- CommentAuthorspitters
- CommentTimeMar 6th 2014
- PermaLink
Author: spitters
Format: MarkdownItex@Mike every constructively *definable* (total) function to the Dedekind real is uniformly continuous. So, it seems silly to constructively define a general theory without examples. Bishop considers functions with a "full" domain. This works well. Your suggestion to consider functions to the lower reals, lower continuous functions, may work well. For some reason, these lower reals did not show up in the Bishop-style literature until recently.

@Mike every constructively definable (total) function to the Dedekind real is uniformly continuous. So, it seems silly to constructively define a general theory without examples. Bishop considers functions with a “full” domain. This works well.

Your suggestion to consider functions to the lower reals, lower continuous functions, may work well. For some reason, these lower reals did not show up in the Bishop-style literature until recently.
- CommentRowNumber21.
- CommentAuthorMike Shulman
- CommentTimeMar 6th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItex*A priori* I don't see why a lack of *definable* examples would be a problem. The point is to have a general theory which is well-behaved. On one hand, classically, the theory of Henstock integration is well-behaved because we have interesting discontinuous gauges. On the other hand, if we assume that all functions are continuous, then the theory of Henstock integration is well-behaved because it coincides with Riemann integration and all functions are Riemann integrable. So if we could do Henstock integration constructively, regardless of whether we had constructively *definable* interesting gauges, then it would encompass both of these cases. However, the fact that one naturally arising example is only a function to the lower reals does suggest that that's a better way to go. Here's another classically important example of a gauge: for $c\in [a,b]$ define $$ \delta(x) = \begin{cases} 1 &\quad x=c\\ \textstyle\frac{1}{2}|x-c| &\quad x\neq c \end{cases} $$ The important property of this is that in any $\delta$-fine partition, the point $c$ must be the tag of the sub-interval which contains it. Now the definition above obviously uses LEM, and is discontinuous, but it is lower semicontinuous and indeed we can define it as a function to the lower reals: $$ \delta(x) = \sup \Big(\Big\{\textstyle\frac{1}{2}|x-c|\Big\} \cup \big\{ 1 | x=c\big \} \Big). $$ Moreover, I think this has the desired property that in any $\delta$-fine partition, the point $c$ must be the tag of the sub-interval which contains it, essentially because the [[apartness]] on the reals is tight. What's not clear to me, however, is whether there is any constructive sense in which this function $\delta$ is "positive".

A priori I don’t see why a lack of definable examples would be a problem. The point is to have a general theory which is well-behaved. On one hand, classically, the theory of Henstock integration is well-behaved because we have interesting discontinuous gauges. On the other hand, if we assume that all functions are continuous, then the theory of Henstock integration is well-behaved because it coincides with Riemann integration and all functions are Riemann integrable. So if we could do Henstock integration constructively, regardless of whether we had constructively definable interesting gauges, then it would encompass both of these cases.

However, the fact that one naturally arising example is only a function to the lower reals does suggest that that’s a better way to go. Here’s another classically important example of a gauge: for $c\in [a,b]$ define
$\delta(x) = \begin{cases} 1 &\quad x=c\\ \textstyle\frac{1}{2}|x-c| &\quad x\neq c \end{cases}$
The important property of this is that in any $\delta$ -fine partition, the point $c$ must be the tag of the sub-interval which contains it. Now the definition above obviously uses LEM, and is discontinuous, but it is lower semicontinuous and indeed we can define it as a function to the lower reals:
$\delta(x) = \sup \Big(\Big\{\textstyle\frac{1}{2}|x-c|\Big\} \cup \big\{ 1 | x=c\big \} \Big).$
Moreover, I think this has the desired property that in any $\delta$ -fine partition, the point $c$ must be the tag of the sub-interval which contains it, essentially because the apartness on the reals is tight. What’s not clear to me, however, is whether there is any constructive sense in which this function $\delta$ is “positive”.
- CommentRowNumber22.
- CommentAuthorspitters
- CommentTimeMar 7th 2014
- (edited Mar 7th 2014)
- PermaLink
Author: spitters
Format: MarkdownItex$\delta$ is positive on an open dense set, a full set (measure 1), and forall $x$, not $\delta(x)\leq 0$.

$\delta$ is positive on an open dense set, a full set (measure 1), and forall $x$ , not $\delta(x)\leq 0$ .
- CommentRowNumber23.
- CommentAuthorMike Shulman
- CommentTimeMar 7th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexThanks! The first two are weaker than positivity even in classical logic; can we weaken the requirement of positivity for the gauge in that way? We are probably still going to have the problem of the *existence* of $\delta$-fine tagged partitions...

Thanks! The first two are weaker than positivity even in classical logic; can we weaken the requirement of positivity for the gauge in that way? We are probably still going to have the problem of the existence of $\delta$ -fine tagged partitions…
- CommentRowNumber24.
- CommentAuthorspitters
- CommentTimeMar 8th 2014
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Author: spitters
Format: MarkdownItexI seem to recall that we need $\delta$ to be strictly positive in $c$ in the example above.

I seem to recall that we need $\delta$ to be strictly positive in $c$ in the example above.
- CommentRowNumber25.
- CommentAuthorMike Shulman
- CommentTimeMar 8th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexWhat?

What?
- CommentRowNumber26.
- CommentAuthorTobyBartels
- CommentTimeMar 9th 2014
- PermaLink
Author: TobyBartels
Format: MarkdownItex>IIRC, for the Henstock integral, we do not have: $f$ integrable implies $|f|$ integrable. Indeed, ${|f|}$ is Henstock-integrable iff $f$ is Lebesgue integrable. Using the Henstock integral is like using conditionally convergent infinite series.

IIRC, for the Henstock integral, we do not have: $f$ integrable implies $|f|$ integrable.

Indeed, ${|f|}$ is Henstock-integrable iff $f$ is Lebesgue integrable. Using the Henstock integral is like using conditionally convergent infinite series.
- CommentRowNumber27.
- CommentAuthorTobyBartels
- CommentTimeMar 9th 2014
- (edited Mar 9th 2014)
- PermaLink
Author: TobyBartels
Format: MarkdownItexFor a pointwise constructive theory of the Henstock integral, my intuition is to use both lower reals and the fan theorem. Possibly either of these could be left out.

For a pointwise constructive theory of the Henstock integral, my intuition is to use both lower reals and the fan theorem. Possibly either of these could be left out.
- CommentRowNumber28.
- CommentAuthorMike Shulman
- CommentTimeMar 10th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexI wish we could call it the "fan axiom".

I wish we could call it the “fan axiom”.
- CommentRowNumber29.
- CommentAuthorTobyBartels
- CommentTimeMar 10th 2014
- PermaLink
Author: TobyBartels
Format: MarkdownItexSurely people have done so. That\'s how I think of it: as an optional axiom.

Surely people have done so. That's how I think of it: as an optional axiom.
- CommentRowNumber30.
- CommentAuthorspitters
- CommentTimeMar 10th 2014
- PermaLink
Author: spitters
Format: MarkdownItex"fan principle" seems more common.

“fan principle” seems more common.
- CommentRowNumber31.
- CommentAuthorTobyBartels
- CommentTimeMar 11th 2014
- PermaLink
Author: TobyBartels
Format: MarkdownItexYes, that\'s nicely vague about its status.

Yes, that's nicely vague about its status.
- CommentRowNumber32.
- CommentAuthorspitters
- CommentTimeMar 11th 2014
- PermaLink
Author: spitters
Format: MarkdownItexNot unlike the *principle* of excluded middle. Regarding the applicability to Henstock integration. It is clear that it is needed, but can be avoided in standard ways. Either, following Bishop, by changing the definitions, or by going point-free.

Not unlike the principle of excluded middle.

Regarding the applicability to Henstock integration. It is clear that it is needed, but can be avoided in standard ways. Either, following Bishop, by changing the definitions, or by going point-free.
- CommentRowNumber33.
- CommentAuthorMike Shulman
- CommentTimeMar 11th 2014
- PermaLink
Author: Mike Shulman
Format: MarkdownItexI do not find it at all clear that it can be avoided in these ways.

I do not find it at all clear that it can be avoided in these ways.

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Atrium > Mathematics, Physics & Philosophy: improper integrals