Lebesgue integral

**problem** · October 23rd 2009, 10:18 PM

Suppose $f$ is a non-negative integrable function on $X$ and $F_n=\{ x \in X : f(x)<1/n \}$ .
Show that $lim_{n \rightarrow \infty} \int_{F_n} f = 0$ .

**Moo** · October 24th 2009, 12:24 AM

Hello,

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

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

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

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

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

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

And hence the result...

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

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

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

And hence the result.

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