Wikipedia claims, if $\sigma$-finite the Dominated convergence theorem is still true when pointwise convergence is replaced by convergence in measure, does anyone know where to find a proof of this? Many thanks!

Statement of the theorem:

Let $\mu$ be $\sigma$-finite, $|f_n|\leq g$ and $f_n\rightarrow f$ in measure, then we must have

$\int f_n \rightarrow \int f$ and $\int|f_n-f| \rightarrow 0$

  • $\begingroup$ BTW, since convergence in $L^1$ implies convergence in Measure, we also have the conclusion of DCT with the hyphotesis of convergence in $L^1$ and the sequence of funcionts also dominated in $L^1$ $\endgroup$ Dec 31, 2022 at 21:16

4 Answers 4


I think there is a point of reviving this post because I think this is also a nice proof pointed out to me when I posted this question by Chris Janjigian.

I have seen this question posted numerous times on this site, so I think there is a point of writing this out.

Let us consider the sequence $\int |f_n-f|$. then consider the following, take a subsequence $\int |f_{n_j}-f|$

For $f_{n_j}$, there must exist a sub-subsequence $f_{n_{j_k}}$ such that $f_{n_{j_k}}$ converges to $f(x)$ almost everywhere. (Since $f_n$, hence $f_{n_j}$ converges to $f(x)$ in measure)

It must also be the case $|f_{n_{j_k}}-f| \leq 2g$, we now apply dominated convergence to see that $\int|f_{n_{j_k}}-f| \rightarrow 0$

What we have shown is that, for every subsequence of $\int |f_n-f|$, we have a further subsequence, which converges to 0. Now using the lemma:

If for every subsequence of $x_n$, there exists a sub-subsequence which converges to 0, then $x_n$ converges to 0.

We are done.

The proof of the last lemma can be found Sufficient condition for convergence of a real sequence

  • $\begingroup$ This is an elegant one. $\endgroup$
    – Sam Wong
    Dec 2, 2020 at 14:14

Let $(X,\mathcal B,\mu)$ be a measure space, $\{f_n\}$ a sequence of functions which converges to $f$ in measure, and for almost every $x$ and all $n$, $|f_n(x)|\leqslant g(x)$, where $g$ is integrable. Then $\lVert f_n-f\rVert_{L^1}\to 0$.

Let $A_k:=\{g\gt 1/k\}$; then $A:=\bigcup_k A_k =\{g\neq 0\}$ and $X\setminus A\subset\bigcap_n\{f_n=0\}\cap\{f=0\}$. We have for each $k$, $$\int_X|f_n(x)-f(x)|d\mu\leqslant 2\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)+\int_{A_k}|f_n(x)-f(x)|\mathrm d\mu(x).$$ If $\lVert f_n-f\rVert_{L^1}$ doesn't converge to $0$, we can find a $\delta>0$ and a subsequence $\{f_{n'}\}$ such that $\lVert f_{n'}-f\rVert_{L^1}\geqslant 2\delta$. We fix $k$ such that $2\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)\leqslant\delta$ (such a $k$ exists by the dominate convergence theorem, since $\lim_{k\to\infty}\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)= \int_{X\setminus A}|g(x)|\mathrm d\mu(x)$ ). Then $$\delta\leqslant \int_{A_k}|f_{n'}(x)-f(x)|\mathrm d\mu(x).$$ Now, as $A_k$ has a finite measure, we can extract a subsequence $\{f_{n''}\}$ of $\{f_{n'}\}$ which converges almost everywhere on $A_k$. Applying the classical dominated convergence theorem to this sequence, we get a contradiction.

  • $\begingroup$ Hi Davide, I don't understand two points in your proof: when you said We fix $k$ such that $2\int_{X\setminus A_k}|g(x)|d\mu(x)\leq\delta$, what gurantees the existence of such a $k$? Also, since $A_k$ has finite measure, why can we then extract a subsequence $\{f_{n''}\}$ of $\{f_{n'}\}$ which converges a.e. on $A_k$? I'm probably missing something quite important here, and thanks for your help! $\endgroup$ Oct 5, 2012 at 12:03
  • 2
    $\begingroup$ For the first point, I apply monotone convergence theorem. For the second one, we use the fact that one a finite measure space, if a sequence converges in measure, we can extract a subsequence which converges almost everywhere. $\endgroup$ Oct 5, 2012 at 12:11
  • $\begingroup$ Thanks! I finally understand it now. $\endgroup$ Oct 5, 2012 at 12:18
  • 7
    $\begingroup$ Davide, I think in a general measure space (not neccessarily finite) if we have a sequence of functions converge in measure, then we can always extract a subsequence that converges almost everywhere. $\endgroup$ Dec 9, 2012 at 23:37
  • 4
    $\begingroup$ It could be useful to observe that the assumption of $\sigma$-finiteness is not actually needed. Since every integrable function vanishes outside of a $\sigma$-finite set, we know $g \equiv 0$ outside of $X' =\bigcup_n A_n$ with each $A_n$ of finite measure. Because of $|f_n| \leq g$ almost everywhere, we also get $f_n \equiv 0$ outside of $X'$ (maybe after adding a set of measure zero). Now, we can restrict to $X'$, apply your proof and thus get the claim for the whole space. $\endgroup$
    – PhoemueX
    May 16, 2015 at 20:53

I don't know why this question is reopened. However, given none of the proofs really pleasing me, I'll present one.
So just by interchanging the integral signs, we see that:
$$\int |f-f_n| = \int_{0}^{\infty} \underbrace{ \mu\left( |f_n-f|>t\right) }_{=:A_n(t)}dt \text{ (1) }$$ And under the condition of boudedness of $f_n$, we have : $$0 \le A_n(t) \le \mu( 2g >t) := G(t)$$ Note that $$\int_{0}^{\infty} G(t)= \int 2g < \infty$$ Thus the DCT conditions for the first integrals are fulfilled, so :
$$ \lim_{n \rightarrow + \infty} \int |f-f_n| = \lim_{n \rightarrow + \infty} \int_{0}^{\infty} \mu\left( |f_n-f|>t\right) = \int_{0}^{\infty} \lim_{n \rightarrow + \infty} \mu\left( |f_n-f|>t\right) = 0 $$

  • 1
    $\begingroup$ thanks ,Nice proof $\endgroup$
    – yi li
    Nov 12, 2020 at 15:11
  • $\begingroup$ can this b modified when we use henstock variation henstock urzweil integral and convergencein measure for arbitrary not necesarily measurable vector valued hk integrable mapings. $\endgroup$ Mar 10, 2021 at 18:14
  • $\begingroup$ nice pguide me [email protected] $\endgroup$ Mar 10, 2021 at 18:18

Here's another solution. Let $(f_{n_{k}})$ be a subsequence of $(f_{n})$ that converges to $f$ almost everywhere. Then $f\in L^{1}(X, \mu)$. Now let $Y$ be set of finite measure such that $\int_{X\setminus Y}g\:\text{d}\mu < \varepsilon/8$. Choose $\delta > 0$ such that if $S$ is any measurable set satisfying $\mu(S) < \delta$ then $\int_{S}g\:\text{d}\mu < \varepsilon/8$. Choose $N$ large enough such that if $n\geq N$ then $\mu(A_{n\varepsilon}) < \delta$ where \begin{equation*} A_{n\varepsilon} = \{x\in X : |f_{n}(x)-f(x)| > \varepsilon/2\mu(Y)\}. \end{equation*} Then for $n\geq N$ we have \begin{align*} \int |f_{n}-f|\:\text{d}\mu &= \int_{X\setminus Y}|f_{n}-f|\:\text{d}\mu+\int_{Y\cap A_{n\varepsilon}}|f_{n}-f|\:\text{d}\mu+\int_{Y\setminus Y\cap A_{n\varepsilon}}|f_{n}-f|\:\text{d}\mu\\ &\leq 2\int_{X\setminus Y}g\:\text{d}\mu+2\int_{Y\cap A_{n\varepsilon}}g\:\text{d}\mu+\int_{Y\setminus Y\cap A_{n\varepsilon}}|f_{n}-f|\:\text{d}\mu\\ &< 2\cdot\frac{\varepsilon}{8}+2\cdot\frac{\varepsilon}{8}+\frac{\varepsilon}{2\mu(Y)}\cdot\mu(Y). \end{align*}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .