# Monotone convergence theorem. Pointwise and $\mu$-a.e. versions

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?

• For any measurable function $f$, since $\mu(N) =0$, we have that $\int_{X}f(x)\chi_{ N} (x)d\mu= 0$. So, we have $\int_{X}f(x)\chi_{X\setminus N} (x)d\mu=\int_{X}f(x)d\mu$ (in the sense that if one integral exists so does the other and they are equal). If you want to prove this in all detail, you may want to decompose $f$ into $f^+$ and $f^-$. Commented Aug 5, 2020 at 15:27
• @Ramiro, so by existence you mean that that integral is finite, right? I see what do you mean but my question is slightly different.
– RFZ
Commented Aug 5, 2020 at 15:32
• May we assume that it is know that: "For any measurable function $f$, since $\mu(N) =0$, we have that $\int_{X}f(x)\chi_{ N} (x)d\mu= 0$."? Commented Aug 5, 2020 at 15:33
• @Ramiro, yes. I do know that the Lebesgue integral of measurable function over set of zero measure is zero.
– RFZ
Commented Aug 5, 2020 at 15:36
• @Ramiro, could you show the proof of the above statement, please? I'd be happy to see it and I'll appreciate it!
– RFZ
Commented Aug 5, 2020 at 15:42

You wrote:

$$\lim_{n\to \infty}\int_{X}f_n(x)\chi_{X\setminus N} (x)d\mu=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu.$$

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 $$X$$.

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*}

• Since $g$ is non-negative then $\int_{X}g(x)d\mu \in [0,+\infty]$. If you claim that $\int_{X}f(x)d\mu$ exists we need to show that Lebesgue integrals of both $f^{\pm} (x)$ are finite. due to my definition of integrability. I hope that I was clear.
– RFZ
Commented Aug 6, 2020 at 17:20
• @ZFR 1. What part of "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 $\int_X f(x) d\mu= \int_X g(x) d\mu$. " did you not understand? 2. Why you accept that $g$ may have an infinite integral, but does not accept that $f$ may have an infinite integral? 3. Even if you require $f(x)\geq 0$ for all $x \in X$, it does not ensure that $f$ will be integrable, it does not ensure that $f\in L(X)$. Commented Aug 6, 2020 at 17:42

Let us assume that it is known that, for any measurable function $$f$$, if $$N$$ is a measurable set and $$\mu(N) =0$$, then $$\int_{X}f(x)\chi_{ N} (x)d\mu= 0$$. Let us prove the following result.

For any measurable function $$f$$, if $$N$$ is a measurable set and $$\mu(N) =0$$, then:

1. $$\int_{X}f(x)d\mu$$ exists if and only if $$\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$ exists.
2. Moreover, if one of those integrals exists (and so both exist), then $$\int_{X}f(x)d\mu =\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$.

Now, given any measurable function $$f$$, consider its positive part $$f^+$$ and its negative part $$f^-$$. Both $$f^+$$ and $$f^-$$ are positive measurable functions. So we have

\begin{align*} \int_{X}f^+(x)d\mu &= \int_{X}f^+(x)\chi_{X\setminus N} (x)d\mu +\int_{X}f^+(x)\chi_{ N} (x)d\mu\\ &= \int_{X}f^+(x)\chi_{X\setminus N} (x)d\mu + 0\\ & = \int_{X}(f\chi_{X\setminus N})^+ (x)d\mu \end{align*} and

\begin{align*} \int_{X}f^-(x)d\mu &= \int_{X}f^-(x)\chi_{X\setminus N} (x)d\mu +\int_{X}f^-(x)\chi_{ N} (x)d\mu \\ &= \int_{X}f^-(x)\chi_{X\setminus N} (x)d\mu + 0\\ &= \int_{X}(f\chi_{X\setminus N})^- (x)d\mu \end{align*}

Now, the integral of $$f$$ on $$X$$ exists (is defined) if and only if $$\int_{X}f^+(x)d\mu <+\infty$$ or $$\int_{X}f^-(x)d\mu< +\infty$$ which is equivalent to say that $$\int_{X}(f\chi_{X\setminus N})^+ (x)d\mu+\infty$$ or $$\int_{X}(f\chi_{X\setminus N})^- (x)d\mu< +\infty$$ which is equivalent to say that the integral of $$f\chi_{X\setminus N}$$ on $$X$$ exists (is defined).

Morever, if the integral of $$f$$ on $$X$$ exists (and equivalenty, the integral of $$f\chi_{X\setminus N}$$ on $$X$$ exists), we have

\begin{align*} \int_{X}f(x)d\mu &= \int_{X}f^+(x)d\mu-\int_{X}f^-(x)d\mu \\ &=\int_{X}(f\chi_{X\setminus N})^+ (x)d\mu - \int_{X}(f\chi_{X\setminus N})^- (x)d\mu \\ &=\int_{X}(f\chi_{X\setminus N}) (x)d\mu \\ &=\int_{X}f(x)\chi_{X\setminus N} (x)d\mu \end{align*}

So, we have proved that, for any measurable function $$f$$, if $$N$$ is a measurable set and $$\mu(N) =0$$, then $$\int_{X}f(x)d\mu$$ exists if and only if $$\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$ exists. Moreover, if one of those integrals exists (and so both exist), then $$\int_{X}f(x)d\mu =\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$.

Remark 1: There are TWO definitions that are related but different. One definition regards the existence of the integral (which can be infinite) of measurable functions. This definition requires that just $$f^-$$ or $$f^+$$ has finite integral.The second definition is the definition of $$f$$ being Lebesgue integrable, which requires that both $$f^-$$ and $$f^+$$ have finite integrals.

The first definition is more general and the measurable functions that satisfy it are sometimes called "semi-integrable". Of course, if $$f$$ is integrable (second definition) it is automatically semi-integrable (first definition).

Note that the Monotone Convergence Theorem applies to the more general class of semi-integrable functions, in the sense that it does not require (nor conclude) that the functions have finite integral. The functions $$f_n$$ or $$f$$ may have infinite integrals. No assumption is made that $$f_n$$ or $$f$$ are integrable.

Accordingly, the result I have proved above is also for class of semi-integrable functions.

Remark 2: In the question $$f(x) \geqslant 0$$ a. e. It means that there is $$N$$ measurable such that $$\mu(N)=0$$ and $$f(x) \geqslant 0$$ for all $$x\in X\setminus N$$. In particular, $$f(x)\chi_{X\setminus N} (x) \geqslant 0$$, for all $$x\in X$$ and we have that $$\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$ exists.

In my answer, I prove that $$\int_{X}f(x)d\mu$$ exists if and only if $$\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$ exists. Moreover, if one of those integrals exists (and so both exist), then $$\int_{X}f(x)d\mu =\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$.

So, in the case of the question, from the fact that $$\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$ exists we conclude that $$\int_{X}f(x)d\mu$$ exists and that $$\int_{X}f(x)d\mu =\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$$.

• I see that my definition of Lebesgue integrable function is a bit different than yours. I say that $f(x)\in L(X)$ (means that $f(x)$ is Lebesgue integrable on $X$) iff $f^+(x)\in L(X)$ and $f^-(x)\in L(X)$.
– RFZ
Commented Aug 5, 2020 at 18:01
• @ZFR No. It seems you are confusing two concepts. I added a "Remark" at the end of my answer. I hope it clarifies the point. Commented Aug 5, 2020 at 19:02
• @ZFR Please, let me know if you have any further question. Commented Aug 5, 2020 at 20:05
• Actually i do have a question. I am trying to formulate it rigorously. I will write it very soon.
– RFZ
Commented Aug 5, 2020 at 20:45
• @ZFR Remember that, if $f \ge 0$ a.e., then $f^{-}=0$ a.e. Therefore $\int f^{-}=0$, and $\int f$ is well-defined (possibly is infinite). Commented Aug 6, 2020 at 1:14