Prove that for every $\varepsilon>0$ there is a measurable $E$, with $\mu(E)<\infty$, such that $\int_\Omega{f}d\mu \leq\int_E{f}d\mu+\varepsilon$

Question

Let $(\Omega,\mathfrak{F},\mu)$ be a measure space and $f:\Omega\mapsto [0,\infty]$ a non-negative measurable function such that $\int_\Omega{f}d\mu<\infty$; prove that for every $\varepsilon>0$ there is $E\in\mathfrak{F}$, with $\mu(E)<\infty$, such that $\int_\Omega{f}d\mu \leq\int_E{f}d\mu+\varepsilon$

What I'm trying to do to tackle this proof is using monotone convergence theorem; I'm trying to find a sequence $(f_n)_{n=1}^\infty$ such that $f_n:\Omega\mapsto [0,\infty]$ is non-negative and measurable, $f_1\leq f_2\leq f_3\leq\cdots$ and $f_n\to f$.

If I managed to find such a $(f_n)_{n=1}^\infty$ then I'd have, by using monotone convergence theorem,

$$\int_E {f}d\mu=\int_\Omega {f_n}d\mu\to \int_\Omega {f}d\mu$$

So, by limit definition,

For every $\varepsilon>0$ there exists $n_0\in N$ such that for every $n\geq n_0$ it's true that $$0<\int_\Omega {f}d\mu-\int_E {f}d\mu<\varepsilon$$.

The only problem I'm having is finding such a $(f_n)_{n=1}^\infty$

¿Any ideas, help, etc?

Thanks in advance.

Answer 1

Let $E_n=\{\omega\in\Omega\mid f(x)\geq 1/n\}$. By Markov's inequality, $\mu(E_n)$ is finite for all $n>0$ because $f$ is integrable. Now let $f_n=f\chi_{E_n}$ and apply monotone convergence as you suggested.

Answer 2

Notice that $\nu(A):=\int_{A}f(x)d\mu$ defines a measure on $\Omega$ that is absolutely continuous with respect to $\mu$.

Define $E_n:=\{\omega\in\Omega:n> f(x)\geq \frac{1}{n}\}$. Clearly, as $n\to \infty,$ we see $E_n\to \Omega$.

Thus, $\nu(E_n)\to\nu(\Omega)$ as $n\to \infty$ by continuity of measure.

Choose $N$ so large that for all $n \geq N$, we have $0<\nu(\Omega)-\nu (E_n)<\epsilon$.

Set $E:=E_N$ and verify(?) that $\mu(E)<\infty$.

Please check this to make sure I haven't made any egregious mistakes.