Prove that $\mid f(x)\mid \le 1 $ for almost all x.

Question

Suppose $f$ is a real valued $L^1$ function on $\mathbb R$ such that for all measurable sets $E$, we have $$\left| \int_E f(x)\ \mathsf d m \right| \le m(E) $$ where $m$ is Lebesgue measure. Prove that $| f(x)| \le 1$ for almost all x.

I tried to show it by contradiction, and I stuck could not come up with good ideas so I just wondering can anyone provide me hints.

Answer 1

Hint What can you say about $$E_n := \{ x | |f(x)| > 1 +\frac{1}{n} \}$$

Answer 2

As $f\in L^1$, we have the induced signed measure $\nu(E)= \int_E f\ \mathsf dm$, and by hypothesis, $|\nu(E)|\leqslant m(E)$ for all measurable sets $E$. By the Jordan decomposition, $\nu = \nu^+ - \nu^-$ where $\nu^+$ and $\nu^-$ are finite positive measures, and $|\nu|$, the total variation of $\nu$, is defined by $|\nu|=\nu^+ + \nu^-$. So if there were a set of positive measure $E$ on which $f>1$, this would imply $\nu(E)>0$, so $\nu^+(E)>\nu^-(E)$. But $|\nu(E)|\leqslant m(E)$, which yields $\nu^+(E)+\nu^-(E)\leqslant m(E)$, and hence $\nu^+(E) \leqslant m(E)-\nu^-(E)$, a contradiction.