I want to prove that LDCT(Lebesgue Dominated Convergence Theorem) continues to hold if I replace the hypothesis $f_n \to f$ (convergence pointwise) with $f_n\to f$ (convergence in measure): $$\int fd\lambda=\lim_{n\to\infty}\int f_nd\lambda.$$

  • $\begingroup$ This is not exactly an extension, since convergence in measure does not imply convergence almost everywhere (only that there is an almost everywhere convergent subsequence). $\endgroup$
    – tomasz
    Mar 9, 2014 at 11:25

2 Answers 2


Here is another proof:

I'll prove that $\displaystyle a_n:=\int_\Omega f_nd\mu\longrightarrow l:=\int_\Omega fd\mu$ as $n$ tends to $\infty$.

By a very useful fact in Analysis, it's enough to prove that for each subsequence of $\{a_n\}$ like $\{a_{n_k}\}$, there is a subsequence of $\{a_{n_k}\}$ like $\{a_{n_{k_l}}\}$ which converges to $l$.

Now, given a subsequence $\{a_{n_k}\}$, we have $$f_{n_k}\xrightarrow[]{\;\mu\;}f.\tag{I}$$

By this fact, there exists a subsequence of $\{f_{n_k}\}$ like $\{f_{n_{k_l}}\}$ which $$f_{n_{k_l}}\xrightarrow{\;a.e\;}f.\tag{II}$$ (Note that for $\text{(II)}$ we should have assumed that $\Omega$ is of finite measure, however we can get rid of it as @Leo Lerena pointed in the comments.)

Since $\{f_{n_{k_l}}\}$ is dominated by function $g\in\mathcal{L}^1(\Omega,\mu)$, by the original version of $LDCT$, $$\int_\Omega f_{n_{k_l}}\xrightarrow{\;a.e\;}\int_\Omega f.\tag{III}$$ That is : $$a_{n_{k_l}}\xrightarrow{n\rightarrow\infty}l \tag*{$\square$.}$$


The technique of using subsequences, rather than sequences, is one of the most powerful tools of proof !

  • $\begingroup$ Nice. I like your proof. $\endgroup$
    – yoyostein
    Sep 27, 2016 at 4:51
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    $\begingroup$ Working with sub-sub-sequences is very awesome in Analysis :) $\endgroup$ Sep 27, 2016 at 16:29
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    $\begingroup$ As a comment, this works only when $E$ is of finite measure but it can easily be extended to the case when it isn't. $\endgroup$ Apr 30, 2018 at 16:30
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    $\begingroup$ Actually $\text{(II)}$ is true for $\textbf{every}$ measure space. The only tool we use to prove $\text{(II)}$ is the Borel-Cantelli Lemma, which is true for $\textbf{every}$ measure space. c.f. here . $\endgroup$
    – Sam Wong
    Dec 2, 2020 at 13:56
  • $\begingroup$ Why does $\int_\Omega f_{n_{k_l}}\xrightarrow{\;a.e\;}\int_\Omega f$ ? Shouldn't it just be $\int_\Omega f_{n_{k_l}}\rightarrow\int_\Omega f$ as $l\to\infty$? $\endgroup$ Nov 21 at 8:00

Call $(X,\cal F,\mu)$ the involved measure space. Let $g$ integrable such that $|f_n(x)|\leqslant g(x)$ for almost every $x$.

As $g$ is integrable, denote $X':=\{g\neq 0\}=\bigcup_{n\geqslant 1}\{x,|g(x)|>n^{-1}\}$. Then $X'$ with the induced measure is $\sigma$-finite. Applying this version of dominated convergence theorem, we get that $$\int_{X'}fd\mu=\lim_{n\to +\infty}\int_{X'}f_nd\mu.$$ As $X\setminus X'=\{g=0\}\subset \{f=0\}\cup\bigcap_{n\geqslant 1}\{f_n=0\}$, we have $\int_{X'}fd\mu=\int_Xfd\mu$.

So fore each $n$, $\int_{X\setminus X'}fd\mu=\int_{X\setminus X'}f_nd\mu=0$, giving the wanted result.

  • $\begingroup$ Is it true on $R^n$? $\endgroup$
    – 89085731
    Nov 17, 2012 at 22:45
  • $\begingroup$ Yes (and in an arbitrary measured space when we have a non-negative measure). $\endgroup$ Nov 17, 2012 at 22:47

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