Integration of hyperreal functions / Intermediate Value Theorem

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.

-

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 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 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). –  Hurkyl Feb 22 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. –  Hurkyl Feb 22 at 19:39
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. –  Hurkyl Feb 23 at 4:23