# 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). – tomasz Mar 9 '14 at 11:25

## 2 Answers

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 $R^n$? – 89085731 Nov 17 '12 at 22:45
• Yes (and in an arbitrary measured space when we have a non-negative measure). – Davide Giraudo Nov 17 '12 at 22:47

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 : $$a_{n_{k_l}}\xrightarrow{n\rightarrow\infty}l \tag*{\square.}$$

## Note

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

• Nice. I like your proof. – yoyostein Sep 27 '16 at 4:51
• Working with sub-sub-sequences is very awesome in Analysis :) – Fardad Pouran Sep 27 '16 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. – Leo Lerena Apr 30 '18 at 16:30