# Lebesgue integral

• Oct 23rd 2009, 10:18 PM
problem
Lebesgue integral
Suppose $\displaystyle f$ is a non-negative integrable function on $\displaystyle X$ and $\displaystyle F_n=\{ x \in X : f(x)<1/n \}$.
Show that $\displaystyle lim_{n \rightarrow \infty} \int_{F_n} f = 0$.
• Oct 24th 2009, 12:24 AM
Moo
Hello,

Let $\displaystyle G_n=X\backslash F_n=\{x\in X ~:~ f(x)\geq 1/n\}$

We have $\displaystyle \int_X f ~d\mu=\int_{F_n} f ~d\mu+\int_{G_n} f ~ d\mu$ (*)

Let's study $\displaystyle \int_{G_n} f ~ d\mu=\int_X f \cdot\bold{1}_{G_n} ~d\mu$

Since $\displaystyle \left(\tfrac 1n\right)_{n\geq 1}$ is a decreasing sequence, $\displaystyle G_n \subseteq G_{n+1}$ and hence $\displaystyle (f \cdot\bold{1}_{G_n})_{n\geq 1}$ is a non-negative integrable and increasing sequence.

So we can apply Lebesgue monotone convergence theorem, and we have $\displaystyle f \cdot\bold{1}_{G_n} \xrightarrow[]{n\to\infty} f$.

Hence by making the limit as n tends to infinity in (*), we have $\displaystyle \int_X f ~d\mu=\lim_{n\to\infty} \int_{F_n} f ~d\mu+\int_X f ~d\mu$

And hence the result...

Another direct way would've been to use Lebesgue's dominated convergence theorem... :
$\displaystyle \int_{F_n} f~d\mu=\int_X f \cdot\bold{1}_{F_n} ~d\mu$

Note that $\displaystyle f \cdot\bold{1}_{F_n} \leq f$, which is integrable.

And since $\displaystyle \lim_{n\to\infty}F_n=\emptyset$, $\displaystyle f\cdot\bold{1}_{F_n}\xrightarrow[] {n\to\infty} 0$

And hence the result.