Monotone Convergence Theorem for general measure:

Let $(X,\Sigma,\mu)$ be a measure space. Let $f_1, f_2, ...$ be a pointwise non-decreasing sequence of $[0, \infty]$-valued $\Sigma-$measurable functions, i.e. for every $k\ge 1$ and every $x$ in $X$, $$0 \le f_k(x) \le f_{k+1}(x).$$ Next, set the pointwise limit of the sequence ${f_n}$ to be $f$. That is, for every $x$ in $X$, $$f(x) = \lim_{k\to \infty}f_k(x).$$ Then $f$ is $\Sigma-$measurable and $$\lim_{k\to \infty}\int f_k d\mu = \int f d\mu.$$

I've noticed that when it comes to monotone convergence theorem (either Lebesgue or general measure), usually its definition restricts the monotone function sequences to be nonnegative. I'm not sure why the 'nonnegative' is necessary.

  • $\begingroup$ Think of $-|x|/n$ on $\mathbb R$. The sequence converges to 0 but the integral doesn't even though it is increasing. $\endgroup$ – Matt Samuel Feb 9 '16 at 4:46
  • $\begingroup$ @MattSamuel: You are right. so negative function can monotonely converge to zero while its integral may not follow? $\endgroup$ – Bear and bunny Feb 9 '16 at 4:54
  • $\begingroup$ it's just like a huge positive function converging to 0. It is symmetric, you just have to flip your perspective. $\endgroup$ – Matt Samuel Feb 9 '16 at 14:53
  • $\begingroup$ @MattSamuel I cannot see how your example work, the integral is $- \frac{|x|x}{n}$ which converge also to $0$! Could you explain to me, please? $\endgroup$ – The_lost Jul 6 '18 at 11:04
  • $\begingroup$ @the The integral doesn't converge to 0 because it is always $-\infty$. $\endgroup$ – Matt Samuel Jul 6 '18 at 11:36

Well, if $f_k$ could be negative, then its integral might not even be defined. For instance, if $X=\mathbb{R}$ with Lebesgue measure and $f_k(x)=x$ for some $k$, there is no good way to define $\int f_k$ (it should morally be "$\infty-\infty$"). On the other hand, the integral of a nonnegative measurable function can always be defined (though it might be $\infty$).

Even if you require $\int f_k$ to be defined for all $k$, if $\int f_k$ is allowed to be $-\infty$, the result can be false. For instance, let $X=\mathbb{N}$ with counting measure and let $f_k(n)=-1$ if $n>k$ and $0$ if $n\leq k$. Then the $f_k$ are monotone increasing and converge pointwise to the constant function $0$, but $\int f_k=-\infty$ for all $k$.

On the other hand, if you require $\int f_k$ to be defined and $>-\infty$ for all $k$, the result is true. Indeed, you can just replace each $f_k$ by $f_k-f_1$ and use the usual version of the theorem, since all these functions are nonnegative (and the equation $\int f_k=\int f_1+\int (f_k-f_1)$ is guaranteed to make sense and be true since $\int f_1>-\infty$).

  • $\begingroup$ @Eric Wofsey In the last case, if you require $\int f_{k}$ to be defined (not necessarily integrable) and $>-\infty$ for all $k$. Why does make sense consider the $\int f_{k} = \int f_{1} + \int (f_{k} -f_{1})$? since $f_{1} - f_{k} \geq 0 $ we can have that $\int f_{1} - f_{k} = +\infty$. If we have too that $\int f_{1} = +\infty$. so, $\int f_{k} = +\infty -\infty$. $\endgroup$ – Alladin Sep 7 '16 at 13:52
  • 2
    $\begingroup$ @Alladin: $f_k-f_1$ is nonnegative, not nonpositive. $\endgroup$ – Eric Wofsey Sep 7 '16 at 18:55

If they could be negative, then the statement would also have to be true for any sequence $f_n$ where $f_n(x)\geq f_{n+1}(x)$ by just reflecting across $0$. But then consider when $f_n:[0,\infty]\rightarrow R, f_n(x)=\frac{x}{n}$.

In this case, for each $x$, $\lim_n f_n(x)=0$, and thus $f_n \rightarrow f=0$ pointwise, but the integrals are infinite for each finite $n$.


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