1. ## Lebesgue Integral

Suppose $f$ is a non-negative integrable function on $X$.For each $n$,define $E_n= \{ x \in X : f(x)>n \}$.
Show that $lim_{n \rightarrow \infty} m(E_n) =0$.
( $m(A)$ is the measure of $A$.)

I was trying to prove that since $f$ is integrable,then $f < \infty a.e.$.Since $E_n$ is a decreasing sequence of sets,so $lim_{n \rightarrow \infty} m(E_n) = m( \bigcap_{n \in N} E_n)$
Since $f< \infty a.e$ so $\bigcap_{n \in N} E_n = \phi$

so $m( \phi ) = 0$ and hence get the result.
but the problem is I have to assume $m(E_1) < \infty$ to claim the result.I was stuck here.
Can anyone help?Or if this is not the correct approach,can anyone show me the right way?

2. Hello,

What do you think of that ?

I first started by looking at the integral of f over $E_n$ :

$\int_{E_n} f ~d\mu=\int_{\mathbb{R}} f \cdot \bold{1}_{\{f>n\}} ~d\mu$

But in this case, we have $f>n$. Otherwise, it's 0.
So $\int_{E_n} f~d\mu>\int_{\mathbb{R}} n \cdot \bold{1}_{\{f>n\}} ~ d\mu=n\int_{\mathbb{R}} \bold{1}_{\{f>n\}} ~d\mu=n\mu(E_n)$

Now, we know that f is integrable. This means that $\int_{\mathbb{R}}f ~d\mu<\infty$

And since $E_n\subset \mathbb{R}$, we have $\int_{\mathbb{R}}f ~d\mu\geq \int_{E_n}f ~d\mu$

Finally, we get $\infty > \int_{\mathbb{R}} f ~d\mu> n\mu(E_n)$

Now what happens if n goes to infinity ? Unless $\mu(E_n)\to 0$, $n\mu(E_n) \to \infty$...
A simple proof by contradiction can finish it