Fundamental Theorem of Calculus and open intervals

indefinite-integralsordinary differential equationsreal-analysis

I am looking at the following theorem for separable differential equations of first order.

Theorem:

Let $I \subset \mathbb{R}$ be an interval, $\Omega \subset \mathbb{R}$ be open, $g:I \to \mathbb{R}$ and $h:\Omega \to \mathbb{R}$ be continuous and $t_{0} \in I$ as well as $y_{0} \in \Omega$. If $h(y_{0}) \neq 0$, then there exists an interval $J \subset I$ such that the Initial Value Problem

$\left\{\begin{array}{} y'(t)=g(t)h(y(t)), t \in I\\
y(t_{0})=y_{0}\\
\end{array} \right.$

has exactly one solution and it is given by

$u=H^{-1} \circ G$ where $G(t)=\int^{t}_{t_{0}} g(s) ds$ and $H(y)=\int^{y}_{y_{0}} \frac{1}{h(x)} dx$.

However, I do not understand part of the proof where the Fundamental Theorem of Calculus is used. The first step in the proof is to check that $u=H^{-1} \circ G$ makes sense.

Proof (extract):

Since $h$ is continuous and $h(y_{0}) \neq 0$ there
exists an open interval $U \subset \Omega$ such that $h(y) \neq 0$ for
all $t \in U$. Then the FTC implies that $H$ is continuously
differentiable on $U$ and $H(y_{0})=0$.

Why can we use the FTC for functions defined on open intervals? The theorem is stated only for functions on closed intervals $[a,b]$ and more generally integration is always considered for functions on closed intervals. I also know that we can change a function $f$ on finitely many points in $[a,b]$ without changing the integral. But I still don't understand how we can use this to apply the FTC for functions on open or half-open intervals ($g$ is defined on an interval which may be open or half-open as well). My idea was that there must be a way to continuously extend those functions to the endpoints and then use the FTC to obtain an antiderivative. Afterwards, we could simply restrict the function to the original domain again. However, I am not sure if the limit at $a$ or $b$ must necessarily exist given that we only have continuity on the open or half-open interval and nothing else.

Any help is much appreciated. Thanks.

Best Answer

I've thought a bit more about my question and I recalled that there is actually a different version of the FTC that we can use to show that $H$ is an antidervatie of $1/h$.

First note that by the Lebesgue criterion for Riemann integrability we can simply extend $h$ and thus $1/h$ to a closed interval, i.e. let $U=(a,b)$, then we can extend it to $[a,b]$. Of course, this extension is not necessarily continuous at the endpoints $a$ or $b$, but it is still Riemann integrable.

Now we can use the following version of the FTC.

Fundamental Theorem of Calculus (alternative version):

Let $f$ be an integrable function on $[a,b]$. Define

$F(x)=\int^{x}_{a} f(t) dt, x \in [a,b]$.

Then $F$ is continuous on $[a,b]$. Moreover, if $f$ is continuous at $x_{0} \in (a,b)$, then $F$ is differentiable at $x_{0}$ and $F'(x_{0})=f(x_{0})$.

So we can use this to show that $H$ is an antiderivative of $1/h$ on $U$.

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