Tchebyshev inequality for Measure theory.

Question

Suppose $f$ integrable on measurable set $E$ and $f(x)$ $\geq$ $0$ on $E$,

And for any $a \in \mathbb{R}^+$, we define:

$E_a$ = $\{x \in E$ $\vert$ $f(x) > a\}$. (which is measurable by basic principles)

Show that $m(E_a) \leq \frac{1}{a} \int_E f(x)$. ($m$ is Leb Meas)

here is what I have:

Pf:

Since

$a$ $\chi_{E_a}$ $\leq f(x)$ ($\because$ of the fact: $x \in E_a$)

by integrating both sides over $E$ we obtain

$a \int_E \chi_{E_a}$ $\leq$ $\int_E f(x)$

and by definition, we rewrite the LHS and have:

$m(E_a) \leq \frac{1}{a} \int_E f(x)$

does this work?

a classmate said I have to consider an infinite case too? but then

$\int_{E_a} a \chi_{E_a \cap [-n,n]}$ goes off to infinity?

not sure I get this part, is my part correct? thanks in advance!

Answer 1

everything looks fine, you do not have to consider the infinite case since if the integral of f is not finite, then you have the claim trivially. If the integral of the indicator of function over the set you are given is infinte, then by monotonicity of the integral also the integral of f is infinite and again it holds :D