Easy question on derivative in the sense of distribution

Question

I would like help proving this elementary result:

Let $f\in L^{1}_{loc}(a,b)$. Let $x_0 \in (a,b)$ Let $F(x)=\int^{x}_{x_0} f$. Then $F'=f$ in the sense of distributions.

i.e How do I show $<F',\phi>=<f,\phi>$?

All I know is that $<F',\phi> = -<F, \phi'>$. I do not see how the right hand side is equal to $<f,\phi>$

Answer 1

Let $\phi$ be a test function. Let $K$ be a compact interval in $(a,b)$ containing the support of $\phi$. Then $f\in L^1(K)$, as it's locally integrable and $K$ is compact.

We can approach $f$ in $L^1(K)$ by continuous functions $f_k$, e.g. $\lVert f-f_k\rVert\leq k^{-1}$. We have, integrating by parts: $$\int_a^b\int_{x_0}^xf_k(t)dt\phi'(x)dx=-\int_a^bf_k(x)\phi(x)dx.$$ Then take the limit $k\to +\infty$.