Almost complete proof that $\int_A f_n \to \int_A f$

Question

This problem is from Bass. I have one problem with my proof that I have not been able to solve. It might be possible that this is not even the right way to solve it, but this is what I came up with.

Here, $\int$ is equivalent to $\int_\mathbb{R}$

Problem:

Let $(X,\mathcal{A},\mu)$ be a measure space. Suppose you have the functions $f_n$ and $f$ which are integrable and non-negative (for all $n$ in the case of $f_n$). Also, assume that $f_n \to f$ almost everywhere, and $\int f_n \to \int f$. Prove that, for every $A \in \mathcal{A}$.

$$\int_Af_n\mathrm{d}\mu \to \int_A f\mathrm{d} \mu$$

My attempt:

We can rewrite the integral of $f_n$ as

$$\int_A f_n \mathrm{d}\mu=\int f_n \chi_A\mathrm{d}\mu$$ where $\chi_A$ is the indicator function for the set $A$.

For all $n$, define the functions

$$g_n(x):=f_n(x) \chi_A (x)= \begin{cases} f_n(x), & x\in A \\ 0, & x \notin A \end{cases} $$ For each $n$, $g_n(x) \geq 0$, so $|g_n(x)| = g_n(x) \leq f_n(x) $

Now we obtain the limit

$$\lim_{n\to \infty} g_n = \lim_{n\to \infty} [f_n(x) \chi_A(x)] = \chi_A(x) \lim_{n\to \infty}f_n(x) = f(x)\chi_A(x)$$ for all $x$ such that $f_n(x) \to f(x)$.

Since $f_n \to f$ almost everywhere, it follows that $g_n \to g = f \chi_A$ almost everywhere.

Furthermore, by hypothesis, since each $f_n$ is integrable, we get

$$g_n \leq f_n \text{ a.e.}\Rightarrow |g_n| \leq |f_n| \text{ a.e.} \Rightarrow \int |g_n| \mathrm{d}\mu \leq \int |f_n| \mathrm{d}\mu < \infty$$

Therefore, for all $n$ we have that $g_n$ is integrable.

The functions $g_n$ are measurable since both the indicator function and $f_n$ are measurable. Therefore, the multiplication of both functions is also measurable.

Therefore, we have found that $g_n$ are measurable, and that $g_n \to g$ almost everywhere.This means that I have almost all (not pun intended) of the requirements to use the Dominated Convergence Theorem. I am missing finding an integrable function $h: X \to [0, \infty]$ such that, $|g_n(x)| \leq h(x)$ a.e.

The closest I got to this was the previous statement in this answer that $|g_n(x)| = g_n(x) \leq f_n(x)$ a.e.

I don't know how to find a function that absolutely bounds $g_n$ and does not depend on $n$, and, since we did not assume $f_1 \leq f_2 \leq ...$ I can't say that $f_n \leq f$ which would solve this problem. What am I missing?

(For completion, I will finish the proof assuming I found such a function $h$ that lets me use the DCT)

If we invoke the DCT, we have that

$$\lim_{n\to \infty} \int_A f_n \mathrm{d}\mu = \lim_{n\to \infty} \int f_n \chi_A \mathrm{d}\mu = \lim_{n\to\infty} \int g_n \mathrm{d}\mu= \int \lim_{n\to\infty} g_n \mathrm{d}\mu = \int g \mathrm{d}\mu = \int f \chi_A \mathrm{d}\mu = \int_A f \mathrm{d}\mu$$

Thank you!

Answer 1

$|\int_A f_n -\int_A f| \leq \int_X |f_n-f|$. Let us show that $\int_X |f_n-f| \to 0$. $(f-f_n)^{+}\to 0$ almost everywhere and $0 \leq (f-f_n)^{+} \leq f$. By DCT we get $\int_X (f-f_n)^{+}\to 0$. Now $\int_X (f-f_n)^{-}=\int_X (f-f_n)^{+} -\int_X (f-f_n) \to 0-0=0$. Hence $\int_X |f-f_n|=\int_X (f-f_n)^{+}+\int_X (f-f_n)^{-} \to 0$.

Answer 2

\begin{align*} \int f\chi_{A}\leq\liminf\int f_{n}\chi_{A}\leq\limsup\int f_{n}\chi_{A}. \end{align*} While \begin{align*} \limsup\int f_{n}\chi_{A}&=\limsup\left(\int f_{n}-\int f_{n}\chi_{A^{c}}\right)\\ &\leq\limsup\int f_{n}+\limsup\left(-\int f_{n}\chi_{A^{c}}\right)\\ &=\lim\int f_{n}-\liminf\int f_{n}\chi_{A^{c}}\\ &=\int f-\liminf\int f_{n}\chi_{A^{c}}\\ &\leq\int f-\int f\chi_{A^{c}}\\ &=\int f\chi_{A}. \end{align*}