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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?

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  • $\begingroup$ 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 $f^-$. $\endgroup$
    – Ramiro
    Commented Aug 5, 2020 at 15:27
  • $\begingroup$ @Ramiro, so by existence you mean that that integral is finite, right? I see what do you mean but my question is slightly different. $\endgroup$
    – RFZ
    Commented Aug 5, 2020 at 15:32
  • $\begingroup$ May we assume that it is know that: "For any measurable function $f$, since $\mu(N) =0$, we have that $\int_{X}f(x)\chi_{ N} (x)d\mu= 0$."? $\endgroup$
    – Ramiro
    Commented Aug 5, 2020 at 15:33
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    $\begingroup$ @Ramiro, yes. I do know that the Lebesgue integral of measurable function over set of zero measure is zero. $\endgroup$
    – RFZ
    Commented Aug 5, 2020 at 15:36
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    $\begingroup$ @Ramiro, could you show the proof of the above statement, please? I'd be happy to see it and I'll appreciate it! $\endgroup$
    – RFZ
    Commented Aug 5, 2020 at 15:42

2 Answers 2

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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*}

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  • $\begingroup$ 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. $\endgroup$
    – RFZ
    Commented Aug 6, 2020 at 17:20
  • $\begingroup$ @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 $f\in L(X)$. $\endgroup$
    – Ramiro
    Commented Aug 6, 2020 at 17:42
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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 $\int_{X}f(x)d\mu =\int_{X}f(x)\chi_{X\setminus N} (x)d\mu$.

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  • $\begingroup$ 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 $f^-(x)\in L(X)$. $\endgroup$
    – RFZ
    Commented Aug 5, 2020 at 18:01
  • $\begingroup$ @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. $\endgroup$
    – Ramiro
    Commented Aug 5, 2020 at 19:02
  • $\begingroup$ @ZFR Please, let me know if you have any further question. $\endgroup$
    – Ramiro
    Commented Aug 5, 2020 at 20:05
  • $\begingroup$ Actually i do have a question. I am trying to formulate it rigorously. I will write it very soon. $\endgroup$
    – RFZ
    Commented Aug 5, 2020 at 20:45
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    $\begingroup$ @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). $\endgroup$ Commented Aug 6, 2020 at 1:14

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