Completing the proof that $\int f \, d\mu= \lim \int f_n \, d\mu$

Question

The problem and solution is presented as above, but it doesn't seem that the proof is complete. Could someone show in details why the last step implies that $\int f \, d\mu = \lim \int f_n \, d\mu$ please? Thanks.

Answer 1

You are rigt. The proof is incomplete. Here is one way to complete it.

Completing the proof: (remember that $n$ is fixed). We know then $$\int f d\mu \geq \int \varphi d\mu -\epsilon \mu(X)$$ for any simple function $\varphi \leq f_n$. Since $$\int f_n d\mu = \sup \left \{ \int \varphi d\mu \:|\: 0 \leq \varphi \leq f_n \right \}$$ We can conclude $$\int f d\mu \geq \int f_n d\mu -\epsilon \mu(X)$$

Now note that in the argument using $\varphi$ and $\psi$ the functions $f_n$ and $f$ play a completely symmentric role. So, reversing their roles, we can start peeking a simple function $\varphi \leq f$ and we can conclude $$\int f_n d\mu \geq \int f d\mu -\epsilon \mu(X)$$

So we have, since $X$ has finite measure, $$\left |\int f_n d\mu - \int f d\mu \right | \leq \epsilon \mu(X)$$

So for any $\epsilon>0$, there is a $N$ such that if $n\geq N$, we have $|f_n(x)-f(x)|<\epsilon$ for all $x\in X$, and we also have, as we proved, $\left |\int f_n d\mu - \int f d\mu \right | \leq \epsilon \mu(X)$. So $$\int f d\mu = \lim_{n\to\infty} \int f_n d\mu.$$

Remark: There are simple ways to prove the result. For instance:

Simpler proof: By Cor. 2.10, $f\in M$, and as $f_n\geq 0$ for all $n$, we also have $f\geq 0$, and thus $f\in M^+$. Let $\epsilon>0$, then there is a $N$ such that if $n\geq N$, we have $|f_n(x)-f(x)|<\epsilon$ for all $x\in X$, we have $$ f_n - \epsilon <f < f_n+ \epsilon$$ So, if $n\geq N$, we have $$ 0\leq f_n <f+ \epsilon < f_n+ 2\epsilon$$ and so $$ 0\leq \int f_n d\mu \leq \int f d\mu + \epsilon \mu(X) \leq \int f_n d\mu + 2\epsilon \mu(X) $$ which means, since $X$ has finite measure, that $$\left |\int f_n d\mu - \int f d\mu \right | \leq \epsilon \mu(X)$$

So for any $\epsilon>0$, there is a $N$ such that if $n\geq N$, we have $\left |\int f_n d\mu - \int f d\mu \right | \leq \epsilon \mu(X)$. So $$\int f d\mu = \lim_{n\to\infty} \int f_n d\mu.$$