Real Analysis, Folland Proposition 2.16 and Corollary 2.17 Integration of Nonnegative functions

Question

Background Information:

Proposition 2.16 - If $f\in L^+$, then $\int f = 0$ iff $f = 0$ a.e.

Proof - Suppose $f = \sum_{j}a_j\chi_{E_j}$, then $\int f = 0$ iff $a_j = 0$ or $\mu(E_j) = 0$. In general, if $f = 0$ a.e., and $\phi$ is a simple function with $0\leq\phi\leq f$ then $\phi = 0$ a.e. and we have $$\int f = \sup_{\phi\leq f}\int \phi = 0$$ Conversely, suppose $E_n = \{f\geq \frac{1}{n}\}$, then $$\int f \geq \int \frac{1}{n}\chi_{E_n} = \frac{1}{n}\mu(E_n)$$ So, $$\mu(E_n)\leq n\int f = 0$$ then $$\mu(\{f\geq 0\}) = \mu\left(\bigcup_{n}E_n\right) = 0$$

I am pretty sure this proof is correct. My question now pertains to the Corollary of this proposition:

Question:

Corollary 2.17 - If $\{f_n\}\subset L^+$, and $f_n(x)$ increases to $f(x)$ for a.e. $x$, then $\int f = \lim_{n\rightarrow \infty}\int f_n$.

My idea: Let $f = \sum_{1}^{n}a_j\chi_{E_j}$ then $\{E_j\}_{j}$ is a finite disjoint family of measurable functions such that $X = \bigcup_{j}E_j$. We have that $f_n(x)$ increases to $f(x)$ for a.e. $x$ this means that there is a measurable set $F$ such that $x\in F$ with $\mu(F) = 0$. Then $f_n$ increases to $f$ on $X\setminus E$ and then applying the Monotone Convergence theorem $$\int \chi_{X\setminus E}f_n\rightarrow \int \chi_{X\setminus E} f = \int f$$

My apologies if this is a bit messy. Any suggestions is greatly appreciated.

Answer 1

Note: your proof of 2.17 is incorrect, because you assume that an arbitrary element $f \in L^+$ can be written as $f = \sum_{j=1}^{\infty}a_j \chi_{E_j}$, but in fact this is only true of simple functions. Note that nonnegative simple functions are elements of $L^+$, but a general element of $L^+$ is any measurable function from $X$ to $[0,\infty]$.

We wish to prove Corollary 2.17, which is as follows.

Corollary 2.17: If $f_n$ is a sequence of functions in $L^+$, and $f \in L^+$, and $f_n(x)$ increases to $f(x)$ for almost every $x$, then $\int f = \lim \int f_n$.

This is very similar to Theorem 2.14, the Monotone Convergence Theorem, which states:

Theorem 2.14: If $f_n$ is a sequence of functions in $L^+$, and $f_j \leq f_{j+1}$ for all $j$, and $f = \lim f_n (= \sup f_n)$, then $\int f = \lim \int f_n$.

The only difference is that 2.14 assumes convergence everywhere, whereas 2.17 relaxes this to convergence almost everywhere. Let's see how to use Theorem 2.17 to prove Corollary 2.17.

We will also use Proposition 2.16, which states:

Proposition 2.16: If $f \in L^+$, then $\int f = 0$ iff $f = 0$ almost everywhere.

Proof of Corollary 2.17

We are given that $f_n \in L^+$ and $f \in L^+$, and $f_n(x)$ increases to $f(x)$ for almost every $x$. So, let $E$ denote the set of those $x$ for which $f_n(x)$ does not increase to $f(x)$. Then $E$ has measure zero (by definition of "almost everywhere"), and $f_n(x)$ increases to $f(x)$ for every element of $X \setminus E$. Therefore, $$\int_{X \setminus E}f = \int_X f \chi_{X \setminus E} = \lim \int_X f_n \chi_{X \setminus E} = \lim \int_{X \setminus E}f_n\quad\quad(*)$$ where the middle equality holds by the Monotone Convergence Theorem (2.14) since $f_n \chi_{X \setminus E}(x)$ increases to $f \chi_{X \setminus E}(x)$ for every $x \in X$.

Now, note that $\int_E f = \int_X f \chi_E = 0$ by Proposition 2.16 (since $f \chi_E = 0$ almost everywhere). Similarly, $\int_E f_n = 0$. Equation $(*)$ remains true if we add zero to both sides, so: $$\begin{aligned} \int_X f &= \underbrace{\int_{E} f}_{=0} + \int_{X \setminus E}f \\ &= \lim\underbrace{\int_{E}f_n}_{=0} + \lim\int_{X \setminus E}f_n \\ &= \lim\left(\int_{E}f_n + \int_{X \setminus E}f_n \right) \\ &= \lim \int_X f_n \\ \end{aligned} $$