An integrable function is finite a.e. Let $f:\mathbb R\longrightarrow \mathbb R$ an integrable function and let $F(x)=\int_{-\infty }^x f(t)dt$.
1) Show that $f$ is finite a.e.
2) Show that $F$ is uniformly continuous.
What I tried:
1) I have to show that $m\{x\mid |f(x)|=\infty \}=0$. Let $E=\{x\mid |f(x)|=\infty \}$ and suppose that $m(E)>0$. In particular, $$\int|f|=\int_{\mathbb R\backslash E}|f|+\int_E|f|=\infty $$
which is a contradiction with $f$ integrable.
1) Is it correct ? It sounds to have something wrong.
2) Let $E_k=\{x\mid |f(x)|>k\}$. Then $E_{k+1}\subset E_k$ and $$\{x\mid |f(x)|=\infty \}=\bigcap_{k\in\mathbb N}E_k.$$
Since $|f|\chi_{E_k}\leq |f|$ and that $|f|\in L^1(\mathbb R)$ by dominated convergence, $$\lim_{k\to\infty }\int |f|\chi_{E_k}=\int\lim_{n\to\infty }|f|\chi_{E_k}=\int |f|\chi_E$$
Since $\chi_E=0$ a.e., $$\int |f|\chi_E=0$$
almost every where. 
2) Is it correct ? There is something strange since on $E$, we have $f=\infty $, therefore it looks to have an indetermination
For the rest it's fine. Notice that for 2), I'm trying to prove that $$\lim_{k\to\infty }\int _{E_k}f=0.$$

For 1) it's almost correct. Since $f$ is integrable on $E$, then $f$ is integrable on every measurable subset $A\subset \mathbb R$. In particuluar, If $m(E)>0$, then $$\int_E |f|=\infty $$ which is a contradiction with the fact that $f$ is integrable on $\mathbb R$.
2) Yes it's correct. Indeed,
$$|f|\chi_E=0\ \ \text{a.e.}\implies \int |f|\chi_E=0.$$
