Let $f \in L^1(\mathbb{R})$ and define
$$h(x)=\int_{-\infty}^{x} f(y)dy$$
Prove that $h'=f$ where $h'$ is the weakly derivative of $h$.
$\textbf{Definition:}$ Let $f\in L^{1}(\mathbb{R})$, the weakly derivative $g$ (if exists) satisfies:
$$ \int_{\mathbb{R}} f(t)\varphi'(t)dt = -\int_{\mathbb{R}}g(t)\varphi(t)dt$$
for all $\varphi \in C^{\infty}_0$. (Infinitely differentiable functions with compact support)
$\textbf{My attempt:}$ Let $\varphi \in C^{\infty}_0$, suppose that $\text{supp}(\varphi) \subseteq [a,b]$ with $\varphi(a)=\varphi(b)=0$. Then we have:
$$ \int_{\mathbb{R}} \varphi'(x)h(x)dx = \int_{a}^{b} \varphi'(x)h(x)dx = \int_{a}^{b} \varphi'(x) \left( \int_{-\infty}^x f(y)dy \right) dx $$
$$ \int_{\mathbb{R}} \varphi'(x)h(x)dx = \int_{a}^{b} \varphi'(x) \left( \lim_{x_0 \to -\infty} \int_{x_0}^x f(y)dy \right) dx = \lim_{x_0 \to -\infty} \int_{a}^{b} \varphi'(x) \left( \int_{x_0}^x f(y)dy \right) dx $$
Operating the integrals:
$$ \int_{a}^{b} \varphi'(x) \left( \int_{x_0}^x f(y)dy \right) dx = -\int_{a}^{x_0} \varphi'(x) \left( \int_{x}^{x_0} f(y)dy \right) dx + \int_{x_0}^{b} \varphi'(x) \left( \int_{x_0}^x f(y)dy \right) dx $$
By Fubini's theorem:
$$ \int_{a}^{b} \varphi'(x) \left( \int_{x_0}^x f(y)dy \right) dx = -\int_{a}^{x_0} f(y)dy \left( \int_{a}^{y} \varphi'(x)dx \right) + \int_{x_0}^{b} f(y)dy \left( \int_{y}^{b} \varphi'(x)dx \right) $$
$$ \int_{a}^{b} \varphi'(x) \left( \int_{x_0}^x f(y)dy \right) dx = - \int_{a}^{b} f(y)\varphi(y)dy$$
So:
$$ \int_{\mathbb{R}} \varphi'h = \lim_{x_0 \to -\infty} \int_{a}^{b} \varphi'(x) \left( \int_{x_0}^x f(y)dy \right) dx = - \int_{a}^{b} f(y)\varphi(y)dy = - \int_{\mathbb{R}} f \varphi $$
Then we have $h' = f$. My proof is correct?
$h$ is bounded and absolutely continuous so $h\phi$ is absolutely continuous. [Because $\phi$ is also bounded and absolutely continuous]. Hence $h(y)\phi(y)-h(x)\phi(x)=\int_x^{y} (h\phi)'=\int_x^{y} (h\phi'+f\phi)$ since $h'=f$ almost everywhere. It follows by letting $y \to \infty$ and $x \to -\infty$ that $\int_{-\infty}^{\infty} (h\phi'+f\phi)=0$ which is what we have to prove.