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

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!

\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*}
• I really appreciate your answers, but is it possible to find the function $h$ I was mentioning or is my method just not going to work? Oct 9, 2019 at 5:51
• No I don't think so, because for $A=X$, $g_{n}$ is just $f_{n}$, surely there is no need to have such upper bound. Oct 9, 2019 at 5:54
• Thank you! Two more things. Why is $\int f \chi_a \leq \lim \inf \int f_n \chi_A$ true? Also, is it possible to prove this using the Monotone Convergence Theorem? Oct 9, 2019 at 6:59
$$|\int_A f_n -\int_A f| \leq \int_X |f_n-f|$$. 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 $$\int_X |f-f_n|=\int_X (f-f_n)^{+}+\int_X (f-f_n)^{-} \to 0$$.
• I really appreciate your answers, but is it possible to find the function $h$ I was mentioning or is my method just not going to work? Oct 9, 2019 at 5:52
• @TheBosco There is no dominating function for $g_n$ so your approach does not work. Oct 9, 2019 at 5:56