Monotone convergence theorem (pointwise version). Let $(X,\mathcal{M},\mu)$ be a measure space and
$\{f_n(x)\}_{n=1}^{\infty}$ be a sequence of measurable functions on
$X$ such that $0\leq f_1(x)\leq f_2(x)\leq \dots$ for all $x\in X$.
Let $f_n(x)\to f(x)$ as $n\to \infty$ for all $x\in X$. Then $$\lim
_{n\to \infty}\int_{X}f_n(x)d\mu=\int_{X}f(x)d\mu.$$Remark: The measurability of $f(x)$ on $X$ follows immediately because $f(x)$ is the pointwise limit of measurable functions on $X$.
I reviewed about 5-6 books on measure theory and noticed that we can slightly change the theorem. More precisely
Monotone convergence theorem ($\mu$-a.e. version). Let $(X,\mathcal{M},\mu)$ be a measure space and
$f(x),f_1(x),f_2(x),\dots$ are measurable functions on $X$ such that
$0\leq f_1(x)\leq f_2(x)\leq \dots$ a.e. on $X$. Let $f_n(x)\to f(x)$
as $n\to \infty$ a.e. on $X$. Then $$\lim \limits_{n\to
\infty}\int_{X}f_n(x)d\mu=\int_{X}f(x)d\mu.$$
I know the proof of the pointwise version. I am trying to prove the $\mu$-a.e. version.
Let $N_1=\{x\in X: \text{monotonicity of} \ f_n(x) \ \text{fails}\}$ and $N_2=\{x\in X:f_n(x)\nrightarrow f(x)\}$ then $N_1,N_2\in \mathcal{M}$ with $\mu(N_1)=\mu(N_2)=0.$ Let $N:=N_1\cup N_2$ then $\mu(N)=0$.
Consider the sequence $\varphi_n(x)=f_n(x)\chi_{X\setminus N}(x)$ and $\varphi(x)=f(x)\chi_{X\setminus N} (x)$. We see that $\varphi_n(x), \varphi(x)$ are measurable and $\varphi_1(x)\leq \varphi_2(x)\leq \dots$ for all $x\in X$ and $\varphi_n(x)\to \varphi(x)$ as $n\to \infty$ for all $x\in X$. So we can use the pointwise version of MCT and we obtain the following: $$\lim
_{n\to \infty}\int_{X}f_n(x)\chi_{X\setminus N} (x)d\mu=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu.$$
Since each $f_n(x)$ is nonnegative on $X$ then using linearity and $\mu(N)=0$ we see that $\int_{X}f_n(x)\chi_{X\setminus N} (x)d\mu=\int_{X}f_n(x)d\mu$.
But we cannot use the same reasoning for the $\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$ because $f(x)$ may not be non-negative on $X$. The only we know that $f(x)\geq 0$ a.e. on $X$. But linearity of Lebesgue integral is true for nonnegative functions or integrable functions.
So my question is this:
If we assume that $f(x)\geq 0$ on $X$ then we are done. But what if we do not have this condition? Is it still true?
Can anyone explain it, please?
Best Answer
You wrote:
So, using linearity, you can conclude that
$$\lim_{n\to \infty}\int_{X}f_n(x)(x)d\mu=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$
Your question is how to deal with the right-hand size. Here is a simple solution: Change the values of $f$ on a set of measure zero, to make it non-negative.
In detail:
Since $f(x)\geq 0$ a.e. on $X$, let us define the function $g$ by $g(x) = f(x)$ if $f(x) \geq 0$ and $g(x) = 0$ if $f(x) <0$.
Note that $g$ is non-negative, so $\int_X g(x) d\mu$ exists. Since $g$ is obtained from $f$ by just changing the values of $f$ on a set of measure zero, we have that $\int_X f(x) d\mu$ exists and $\int_X f(x) d\mu= \int_X g(x) d\mu$.
Note also that $f\chi_{X\setminus N}= g\chi_{X\setminus N}$ a.e.. So you can take care of the right-hand side as follows, using the linearity for $g$.
\begin{align*} \lim_{n\to \infty}\int_{X}f_n(x)(x)d\mu &=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu \\ &=\int_{X}g(x)\chi_{X\setminus N} (x)d\mu \\ &=\int_{X}g(x) d\mu \\ &=\int_{X}f(x) d\mu \end{align*}