Let $(f_n)$ be a sequence of non-negative measurable function and $f$ be a non-negative measurable function. It is required to prove that if $(f_n(x))$ increases to $f(x)$ for almost every $x\in \mathbb{R}$ then $\lim_{n\to\infty}\int f_n=\int f$. (By measurable is meant Lebesgue measurable).

The above result is given as a corollary to the so-called monotone convergence theorem which asserts the following.

$(\alpha)$: Let $(f_n)$ be a sequence of non-negative measurable functions such that $f_n(x)\leq f_{n+1}(x)\ \forall n\in\mathbb{Z^+}\ and\ \lim_{n\to\infty}f_n(x)=f(x)\ \forall x\in\mathbb{R}$. Then $\lim_{n\to\infty}\int f_n=\int f$.

I know how to prove this result using Fatou's lemma. But my question is how would one prove it without using Fatou's lemma, i.e. how is it a corollary to $(\alpha)?$ Someone please help. Thanks.

Attempted solution (added later):

Let $E=\{x\in \mathbb{R}|f_n(x)\ \text{increases to}\ f(x) \}$. Then $m(\mathbb{R}-E)=0$. Therefore $\mathbb{R}-E$ and $E$ are measurable. Define $(g_n)$ by $g_n(x)=f_n(x)\ \text{if $x\in E$ and $g_n(x)=0$ if $x\in\mathbb{R-E}$}$. Then $(g_n)$ is a sequence of measurable functions that satisfies the condition in $(\alpha)$. Therefore $\int f=\lim_{n\to\infty}\int g_n=lim_{n\to\infty}\int f_n$. $\blacksquare$

Is the above argument alright?


1 Answer 1


Define $g_n=f_n$ a.e. and $g=f$ a.e such that the conditions of the theorem are true for $g_n$. For example, define $g_n$ and $g$ to be 0 for values $x$ where $f_n$ is not increasing.

Then \begin{equation} \lim_{n\rightarrow \infty} \int f_n = \lim_{n\rightarrow \infty} \int g_n = \int g = \int f \end{equation}

  • $\begingroup$ Thanks. Could you please check the "added later" part of my post? $\endgroup$
    – Janitha357
    May 3, 2017 at 17:22
  • 1
    $\begingroup$ two comments: 1)use backslash for set subtraction. 2) also define the limit function $g$. With these two changes, I think it will be good $\endgroup$
    – Dunham
    May 3, 2017 at 20:12

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