Measure and integration(In specific about $L^1$) The exercise consists of the following:
Let $ \left (X, \mathcal{M}, \mu \right) $ be a measure space and $ f \in L^1 (\mu) $.
If $A \in \mathcal{M}$ is such that $0< \mu(A) < \infty$, show that $|f(x)| \le \frac{1}{\mu(A)} \int_A |f|\,d\mu$ for some $x \in A $.
What ideas can you give me to solve it?
I can't really think of anything.
 A: Hint: suppose that $|f(x)| > \frac{1}{\mu(A)}\int_A |f|\,d\mu$ for a.e. $x$. What happens if you integrate both sides over $A$? Do you see why it is important that $0 < \mu(A) < \infty$ for this argument?
A: The key is to notice that if $g\geq0$ is a integrable with respect to a measure $\mu$,  and $A=\{g>0\}$ has positive measure (i.e. $\mu(A)>0$), then it must be the case that $\int g\,d\mu>0$.
To see that, notie that $\{g>0\}=\bigcup_n\{g>\frac1n\}$. Thus, there is $n_0$ such that $\mu\big(\{g>\frac{1}{n_0}\}\big)>0$. Consequently
$$\int g\,d\mu\geq\int_{\{g>\tfrac{1}{n_0}\}}g\,d\mu\geq\frac{1}{n_0}\mu\big(\{g>\tfrac{1}{n_0}\}\big)>0$$

In your posting, If there is no point $x\in A$ for which $|f(x)|\leq \frac{1}{\mu(A)}\int_A |f|\,d\mu=\alpha_A$, then it means that for all $x\in A$,
$$|f(x)|>\frac{1}{\mu(A)}\int_A\,|f|\,d\mu=\alpha_A$$
Define $g(x):=\big(|f(x)|-\alpha_A\big)\mathbb{1}_A(x)$ (where $\mathbb{1}_A(x)=1$ if $x\in A$ and $0$ otherwise). As $A=\{g>0\}$ and $\mu(A)>0$ by assumption, the statement above implies that
$$0<\int g\,d\mu=\Big(\int_A|f|\,d\mu\Big)-\mu(A)\alpha_A=0$$
which is absurd. The conclusion is that there must be $x\in A$ for which $|f(x)|\leq\alpha_A$.

Obs: Notice that a  similar argument shows that in fact, there exists $x'\in A$ such that $ |f(x')|\geq \alpha_A$.
