Given a distribution $T \in D'(\mathbb{R})$ such that the distributional derivative $\partial T \in L^1\_{loc}(\mathbb{R})$. Can one deduce that $T \in L^1\_{loc}(\mathbb{R})$ as well? Or can anyone give me an example where $T \notin L^1\_{loc}(\mathbb{R})$?



Distributional derivative is locally integrable - then the distribution as well?
Let $T$ be your distribution. By hypothesis 
$$
T'(\phi)=-T(\phi')=\int\_\mathbb R f\phi'dx,
$$
where $\phi$ is a test function and $f\in L^1\_{\text{loc}}$.


Fix $x\_0\in\mathbb R$ and define 
$$
F(x):=\int\_{x\_0}^x f(x)dx.
$$
Then $F$ is a continuous function, which is almost everywhere differentiable; moreover, if $F'(x)$ is defined at some point $x$, then it equals $f(x)$, so that $F'$ and $f$ define the same element in $L^1\_{\text{loc}}$.


Then, the distribution $T$ is given by integration against $F$ up to some additive constant. Indeed, by integration by parts one obtains
$$
\int\_\mathbb R F\phi' dx=-\int\_\mathbb R F'\phi dx=-\int\_\mathbb R f\phi'dx=T(\phi')
$$
This shows that the distribution $T\_F$ defined by $F$ equals $T$ on every test function which is the derivative of a test function (the point is that *a priori* a primitive of a test function is no longer a test function in general). To conclude you just need the following elementary lemma:


**Lemma.** Let $S$ be a distribution such that $S'=0$. Then, $S(\bullet)=\alpha\int\_\mathbb R\bullet dx$, for some $\alpha\in\mathbb R$.


*Proof.* Let $\alpha:=S(\phi\_0)$, where $\phi\_0$ is a test function such that $\int\_\mathbb R\phi\_0 dx=1$. Let $\phi$ be any test function and write it as $\phi=(\phi-c\phi\_0)+c\phi\_0$, where $c=\int\_\mathbb R\phi dx$. Then, $\int\_\mathbb R(\phi-c\phi\_0)dx=0$ so that $\phi-c\phi\_0$ admits a primitive which is actually a test function, for instance 
$$
\Phi\_c:=\int\_{-\infty}^x(\phi(t)-c\phi\_0(t))dt.
$$
Therefore, $S(\phi-c\phi\_0)=S(\Phi\_c')=-S'(\Phi\_c)=0$. But then,
$$
S(\phi)=c S(\phi\_0)=\alpha\int\_\mathbb R\phi dx.\qquad\qquad\square
$$
Thus, since $(T\_F-T)'(\phi)=T(\phi')-T\_F(\phi')=0$ for every test function $\phi$, you have that $T=T\_F+\alpha\int\_\mathbb R\bullet dx$, for some $\alpha\in\mathbb R$.


Therefore, $T=T\_{F+\alpha}$ and $T$ is the distribution associated to the continuous almost everywhere differentiable function $F+\alpha$.


