# Measure Convergence Version of Lebesgue Dominated Convergence Theorem

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): $$\int fd\lambda=\lim_{n\to\infty}\int f_nd\lambda.$$

• 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). Commented Mar 9, 2014 at 11:25

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 $$LDCT$$, $$\int_\Omega f_{n_{k_l}}\xrightarrow{\;\;}\int_\Omega f.\tag{III}$$ That is : $$a_{n_{k_l}}\xrightarrow{n\rightarrow\infty}l \tag*{\square.}$$

## Note

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

• Working with sub-sub-sequences is very awesome in Analysis :) Commented Sep 27, 2016 at 16:29
• 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. Commented Apr 30, 2018 at 16:30
• 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 . Commented Dec 2, 2020 at 13:56
• 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 $l\to\infty$? Commented Nov 21, 2023 at 8:00
• @nolemonnomelon You're right. Thanks. Commented Jan 30 at 12:54

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.

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