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Here's a statement on hyperreal function I've been trying to prove (I came up with it but I think it is true):

Suppose $f(x)$ is a continuous real-valued function and $h(x)$ is a continuous hyperreal-valued function such that $f(x) \approx h(x)$ for all $x$ (even infinite).

If $\int_a^\infty f(x)\,dx$ and $\int_a^\infty h(x)\,dx$ exist and are finite, then $ \int_a^\infty f(x)\,dx = \int_a^\infty h(x)\,dx$.

Note: $\int_a^\infty g(x)\,dx = st\left(\sum_a^Ng(x)\,dx\right)$ for any infinite $N$.

Using notions of standard analysis, I can show that $\int_a^b f(x)\,dx - \int_a^b h(x)\,dx=0$ for all real $b$ and therefore remains $0$ as $b$ grows.

However, I am unable to use nonstandard analysis on this, that is, to show $\sum_a^N f(x)\,dx \approx \sum_a^N h(x)\,dx$ for all positive infinite $N$. Just because the statement holds for all real $b$, I can't see why it must hold for infinite hyperreals and so I feel like my proof is not rigorous enough. The transfer principle will not apply since $h(x)$ does not have to be a natural extension of any real-valued function.

Even more, the statement implies that if $h(x) \approx 0$ and $\sum_0^N h(x)\,dx \approx \sum_0^{M} h(x)\,dx \approx r \in \mathbb{R}$ for all $N,M$, then $r= 0$. So we can't have a sort of infinite series of infinitesimals that equals $1$. This seems surprising to me.

Edit:

Would it be valid to argue that $$\delta(b) = st\left(\sum_a^b f(x)\,dx - \sum_a^b h(x)\,dx\right) = \begin{cases} 0, & b \text{ is finite} \\ r \neq 0, &b \text{ is infinite} \end{cases}$$ is impossible since $\delta(b)$ is a continuous function? $\delta(b)$ is constant for infinite $b$ because the integrals exist and are finite and thus $\delta(N)=\delta(M)$ for any infinite $N, M$.

Edit 2:

Justification of this approach has brought me to another problem altogether. The function $l(x)=st(x)$ is defined on all finite hyperreals and continuous but does not satisfy the Intermediate Value Theorem (IVT) in ${}^*\mathbb{R}$. For example, $l(0)=0$ and $l(1)=1$ but there is no $x \in [0,1]$ such that $l(x)=\epsilon$ for any infinitesimal $\epsilon$. However, there is an $x$ such that $l(x) \approx \epsilon$. In this second sense, $l(x)$ passes the IVT but $\delta(b)$ does not. I still need to work out the details to understand why I had to reword the theorem for hyperreal-valued functions.

Edit 3:

I think I've finally resolved the matter. I tried to see if the hyperreal number line is complete, and as I suspected, the nonstandard metric is defined in terms of $\approx$ or "halos." So instead of points converging, it is the halos that do. That is why I had to reword the Intermediate Value Theorem to use $\approx$ instead of $=$. So yes, $\delta(b)$ can only be continuous if it is $0$ everywhere.

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1 Answers1

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I suppose you mean $f(x)$ is a continuous standard function, and $h(x)$ is a continuous (non-standard) function.

A serious problem with your reasoning is that identities like

$$ \lim_{x \to +\infty} g(x) = \mathop{\text{st}} g(N) $$

(when the r.h.s. has the same value for all infinite $N$) only hold when $g(x)$ is a standard function. Trying to apply the same formula to compute things involving non-standard functions will give nonsensical results, which is the cause of the problems you're seeing.

Two specific counterexamples to your conjecture are, when

$$ \int_0^\infty f(x) = A > 0$$

and $\epsilon$ is a positive infinitesimal, then we have

$$ \int_0^\infty f(x) + \epsilon\, dx = +\infty$$

and

$$ \int_0^\infty (1 + \epsilon) f(x) \, dx = (1 + \epsilon) A $$

(If you really insist on seeing a finite counterexample that is not infinitesimally close to $A$, just 'truncate' $\epsilon$ suitably. e.g. instead use a function that is $\epsilon$ on $[0, 1/\epsilon]$, zero on $[1/\epsilon + 1, \infty)$, and linear between $1/\epsilon$ and $1 / \epsilon + 1$. The integral will then be infinitesimally greater than $A+1$)

The function $l(x) = \mathop{\text{st}}$ represents an even more severe flaw. It is an external function on the hyperreals. The (internal) notion of continuous/discontinuous for non-standard functions only apply to internal functions, so it would be incorrect to even ask if $l(x)$ is continuous!

One could use the external notion of continuity (as you seem to have done), but it will be very poorly behaved.

(However, IIRC, an internal function is internally continuous if and only if it is externally continuous, so at least you have that)

The intermediate value theorem is

If $h(x)$ is a continuous function and $h(a) < v < h(b)$, then there exists $c \in [a,b]$ such that $h(c) = v$

To apply to non-standard functions, it must be interpreted internally. This means

  • $h(x)$ must be an internal function
  • "continuous" must refer to the internal notion of continuity
  • But your examples don't satisfy the hypothesis, $\int_0^\infty f(x)+\epsilon,dx$ is not finite. – genepeer Feb 22 '13 at 19:11
  • And why can't I study both external and internal functions alike in the context of the hyperreal extension. After all, they are indistinguishable in this context, right? Can't the notions of continuous/discontinuous will be slightly modified to make a coherent study? – genepeer Feb 22 '13 at 19:14
  • @genepeer: The point is that things only behave well internally. For example, the integral operator is defined by the transfer principle, so the expression $$\int_0^\infty h(x) , dx $$ doesn't even make sense if $h(x)$ is external (and the standard integral operator doesn't apply). And those things you can give external meaning to (e.g. continuity) will not behave well analytically (e.g. the IVT does not work for external functions). –  Feb 22 '13 at 19:19
  • I should make a clarification: in standard analysis, there is a set of real functions. We can transfer this set to the non-standard model; while it will be a set of hyperreal functions, it will not be all hyperreal functions. However, it will also contain functions that are not the transfer of standard functions. An internal function is, by definition, a member of this set. An external (hyperreal) function is one that is not. –  Feb 22 '13 at 19:39
  • The easy way to distinguish between internal and external hyperreal functions is whether it is constructed 'internally' or if it references to the standard model. e.g. $h(x) = \epsilon$ is just the constant function corresponding to a number; it is internal. If $\epsilon$ is not a standard number, then $h(x)$ is not a standard function. However the $\mathop{\mathrm{st}}(x)$ explicitly uses knowledge that there is a standard and a non-standard model in its construction; therefore we expect it to be an external function (and it is). –  Feb 22 '13 at 19:43
  • I'm sorry I didn't clarify this while posting the question. I was thinking of defining the integral operator on external functions as the limit of the Riemann sum on the function (just as one does in standard analysis?). So conjecture is: 1) an external hyperreal function $h(x)$ such that $$\int_0^N h(x)dx \text{ is finite, and } \int_N^M h(x)dx \approx 0 \text{ for any infinite }N,M$$ 2) An internal function infinitely close to $h(x)$ (the extension of $f$) with a finite integral as well, then the two integrals are approximately equal. Therefore there standard parts are equal. – genepeer Feb 22 '13 at 21:46
  • This are just ideas I had and I understand that "deep down" they may be inconsistent. So I'd appreciate it if you could clarify the misconceptions I have. – genepeer Feb 22 '13 at 21:50
  • This arose from trying to prove Leibniz's Integral Rule (http://en.wikipedia.org/wiki/Leibniz_integral_rule) using nonstandard analysis. I suspected the theorem but didn't know it had been proven. So I tried using hyperreals to show it, since I've just learned about them, and my current conjecture was crucial to proving the rule. I was not entirely convinced of it, so I asked here. – genepeer Feb 22 '13 at 21:56
  • I expect very many things to go wrong with that external definition. One particular way it goes wrong is that a Riemann sum has finitely many terms; so you can only split $[a,b]$ into finitely many parts, and so it's impossible to take a limit as the size of the largest part goes to zero. –  Feb 23 '13 at 02:55
  • The proof of the Leibniz integral rule should be based on the formula for derivatives of standard functions: $$\frac{d}{dx} f(x) = \text{std} \frac{f(x + \epsilon) - f(x)}{\epsilon} $$ In particular, $$ \frac{d}{dx} \int_{y(x)}^{z(x)} f(x,t) , dt = \text{std}\left( \frac{\int_{y(x+\epsilon)}^{z(x+\epsilon)} f(x + \epsilon) , dx - \int_{y(x)}^{z(x)} f(x,t) , dt}{\epsilon} \right)$$ and then break the numerator into parts the usual way, showing the interesting part $\approx \int_y^z f_x(x,t) , dt$ and the other parts are $\approx 0$. All parts of the integrals are internal functions. –  Feb 23 '13 at 02:59
  • ... and because they are internal, the integral can be defined by transfer. As such, it has all of the usual properties of integrals (so long as you stay internal); e.g. $\int_a^b + \int_b^c = \int_a^c$. Note that this approach only proves it for standard $f,y,z$ (which we assumed at the beginning). Once we've done that, should we desire to do so, the transfer principle would imply it's also true if $f,y,z$ are nonstandard (internal) functions. –  Feb 23 '13 at 03:04
  • Of particular note is that if $f,g$ are internal functions (standard or nonstandard) and $f(x) \approx g(x)$ on the finite interval $[a,b]$, then $$\int_a^b f(x) , dx = \int_a^b g(x) , dx $$ One argument for this is the fact: $$ \left|\int_a^b f(x) - g(x) , dx \right| \leq \int_a^b |f(x) - g(x)| , dx \leq (b-a) \max_{x \in [a,b]} |f(x) - g(x)| $$ –  Feb 23 '13 at 03:10
  • Proving the Leibniz integral rule by switching to Riemann sums is an interesting idea; there's probably some way to get that to work as well. But I don't see immediately how to do so. –  Feb 23 '13 at 03:12
  • The bounds were $[0, \infty]$. That is what I did but then I ended with $\int_0^\infty \frac{f(x+\epsilon,t)-f(x,t)}{\epsilon} ,dx$ but the thing inside the integral was hyperreal-value function that was approx. close to the extension of $f_x(x,t)$ but not the same thing. I don't know if it is internal or not, but was I justified in saying the standard value of their integrals were the same? – genepeer Feb 23 '13 at 03:14
  • I don't know if this changes alot but I used the summation notation. So I had $$\frac{d}{dx}\int_0^\infty f(x,t),dt \approx \ldots = \sum_0^N \frac{f(x+\epsilon, t)-f(x,t)}{\epsilon},dt \approx \sum_0^N f_x(x,t),dt$$ Therefore, $$\frac{d}{dx}\int_0^\infty f(x,t),dt = \int_0^\infty f_x(x,t),dt$$ It is that last step that did not seem justified. – genepeer Feb 23 '13 at 03:34
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    It might be worth asking a new question specifically on the Leibniz integral rule, and proving it with NSA. One very important point is that it's not actually true for improprer integrals! That is, to work for improper integrals you need additional hypotheses. Buck says the conditions you need are that $f(x,y)$ and $f_x(x,y)$ are continuous, and that $$\int_c^\infty f_x(x,y) , dy$$ is uniformly convergent. So, at the very least, to prove this with NSA you'll need to know to express uniform convergence in a useful way. –  Feb 23 '13 at 04:23
  • Thanks! While I was ok with the proof on bounded integrals, the infinite one didn't seem right and I think it's cause I did not consider uniform convergence. Actually, I do not know what that means... yet :) I'll need to read about it before I ask a new question. – genepeer Feb 23 '13 at 04:39
  • Why would integral epsilon dx be infinite? – lalala Jun 03 '21 at 16:48