Let $f\in L_{loc}(\mathbb{R})$, we define the function $g(x)=\int_0^xf(t)dt$. The quesion is to show that the derivative of $g$ is $f$ in the in the distributional sense.
I know the locally integrable function define distributions, but how is $g$ a distribution??
To show that, let $x\in\mathbb{R}$ and $(x_n)_{n\in\mathbb{N^*}}$ a sequence convergent to $x_0$. We show that $(g(x_n))_n$ converges to $g(x_0)$ $$g(x_n)=\int_0^{x_n}f(t)dt=\int_\mathbb{R}\mathbb{1}_{[0;x_n]}(t)f(t)dt$$ the sequence of functions $(1_{[0;x_n]}f)_n$ converges pointwise to $1_{[0;x_0]}f$, and we know that every convegent sequence is bounded, hence, there exist an $M>0$ such that $[0;x_n]\subset[-M;M]$, which yields the domination $$|{1}_{[0;x_n]}(t)f(t)|\leq {1}_{[-M;M]}(t)|f(t)|$$ Notice that the function on the right is integrable. So by the dominated convergence theorem we get the desired result.
2. $g'=f$ in the in the distributional sense.
Let $\varphi \in C_{0}^{\infty}(\mathbb{R})$ a test function and let $A>0$ such that $supp (\varphi) \subset[-A, A]$. By Fubini, we get: $$ \begin{aligned} <g^{\prime}, \varphi>&=-<g, \varphi^{\prime}>=-\int_{-A}^{A}\left(\int_{0}^{x} f(t) \mathrm{d} t\right) \varphi^{\prime}(x) \mathrm{d} x \\ &=-\int_{0}^{A} \int_{0}^{x} f(t) \varphi^{\prime}(x) \mathrm{d} t \mathrm{~d} x+\int_{-A}^{0} \int_{x}^{0} f(t) \varphi^{\prime}(x) \mathrm{d} t \mathrm{~d} x \\ &=-\int_{0}^{A} g(t)\left(\int_{t}^{A} \varphi^{\prime}(x) \mathrm{d} x\right) \mathrm{d} t+\int_{-A}^{0} f(t)\left(\int_{-A}^{t} \varphi^{\prime}(x) \mathrm{d} x\right) \mathrm{d} t \\ &=\int_{0}^{A} f(t) \varphi(t) \mathrm{d} t+\int_{-A}^{0} f(t) \varphi(t) \mathrm{d} t=\int_{\mathbb{R}} f(t) \varphi(t) \mathrm{d} t=<f, \varphi> \end{aligned} $$ Which shows that $g^{\prime}=f$ in the distributional sense.