Let $(f_n)_{n \in \Bbb N}$ be a sequence of measurable functions on $E$, that converges almost everywhere pointwise towards $f$.
Let $(g_n)_{n \in \Bbb N}$ be a sequence of integrable functions on $E$ that converge almost everywhere on $E$ pointwise towards $g$.
Also, suppose that $|f_n| \le g_n$ $\forall n \in \Bbb N$.
I have to show that:$$\lim_{n\to \infty} {\int_E g_n } = \int_E g \Rightarrow \lim_{n\to \infty} {\int_E f_n } = \int_E f$$

I don't really understand how i should show this. I don't see why the right hand side of the relation isn't just allways true due to dominated convergence (i can simply pick one $g_n$). Any ideas or tipps on how to show this and on why it isn't already true because of dominated convergence?

  • $\begingroup$ You can't simply pick one $g_n$, because for $m\neq n$ you don't know if $|f_m|<g_n$ $\endgroup$
    – user160738
    Dec 11, 2016 at 9:02
  • $\begingroup$ Indeed this is a stronger result than dominated convergence. Luckily, you can just copy the proof of dominated converge with some obvious modifications to show this result! $\endgroup$
    – air
    Dec 11, 2016 at 9:04
  • $\begingroup$ I don't understand why to say that $g_{n}$-s converges almost everywhere pointwise. Why just not to say that is converges almost everywhere(without using "pointwise", as it is obvious). Yes you can take on g, $h(x)=sup_{n}{g_{n}(x)}$. What I mean, is that $ |f_{n}(x)| \leq h(x) $ which is convergent almost everywhere as well. $\endgroup$
    – kolobokish
    Dec 11, 2016 at 9:13

2 Answers 2


For the Dominated Convergence to apply, you must have that the $f_n$'s are dominated by a function $\varphi$ whose integral is finite.

Now I added one extra condition to the given because I don't see how it is possible without it. So all in all the conditions become

  • $\lim_{n\to\infty}f_n=f$
  • $\lim_{n\to\infty}g_n=g$
  • $|f_n|\leq g_n$.
  • $\lim_{n\to\infty}\int g_nd\mu=\int g d\mu$
  • $\int g d\mu<\infty$ (extra condition)

Now notice that it is enough to show
$$\lim_{n\to\infty}\int|f_n-f|d\mu=0 $$ The result you want follows immediately from the triangle inequality. To start the proof, define $$\varphi_n=g_n+g-|f_n-f| .$$ Then $\varphi_n$ is positive measurable since
$$|f_n|\leq g_n\implies|f|\leq g\implies\varphi_n\geq g_n+g-(|f_n|+|f|)\geq g_n+g-(g_n+g)=0.$$ Also notice that $\varphi_n\to 2g$ as $n\to\infty$ almost everywhere on $E$. Therefore, we can use Fatou's Lemma to deduce $$\int 2g d\mu=\int2\lim_{n\to\infty}g_nd\mu= \int\lim_{n\to\infty}\varphi_nd\mu= \int\liminf_{n\to\infty}\varphi_nd\mu\leq\liminf_{n\to\infty}\int \varphi_nd\mu,$$ and furthermore, $$\varphi_n\leq g_n+g\implies\int \varphi_nd\mu\leq\int(g_n+g)d\mu\implies \limsup_{n\to\infty}\int \varphi_nd\mu\leq\int 2gd\mu.$$ Hence $$\lim_{n\to\infty}\int \varphi_nd\mu=\int 2g d\mu.$$ Since $$\int (g_n+g) d\mu=\int\varphi_nd\mu+\int |f_n-f|d\mu,$$ then by letting $n\to\infty$ we see that $$\int 2gd\mu=\int 2gd\mu+\lim_{n\to\infty}\int|f_n-f|d\mu.$$ Before you rush and do the cancellation, you need to make sure that $\int2gd\mu$ is finite which is why I needed the extra condition. You are done. Notice that this is very similar to the proof of the Dominated Convergence.

  • 1
    $\begingroup$ And without the condition $\int g\,d\mu < +\infty$, the conclusion need not hold, let $f_n = \chi_{[n,n+1]}$ and $g_n = \chi_{[0,2^n]}$ for an example. $\endgroup$ Dec 11, 2016 at 11:30
  • 1
    $\begingroup$ You can shorten the last part using $$\liminf_{n\to\infty} \int \varphi_n\,d\mu = 2\int g\,d\mu - \limsup_{n\to\infty} \int \lvert f_n - f\rvert\,d\mu.$$ $\endgroup$ Dec 11, 2016 at 11:34

In fact, we can prove the following more generalized conclusion.

Theorem (The extended dominated convergence theorem or $E D C T) . \quad$ Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f_{n}, g_{n}: \Omega \rightarrow \mathbb{R}$ be $\langle\mathcal{F}, \mathbb{R}\rangle$-measurable functions such that $\left|f_{n}\right| \leq g_{n}$ a.e. ( $\left.\mu\right)$ for all $n \geq 1$. Suppose that (i) $g_{n} \rightarrow g$ a.e. $(\mu)$ and $f_{n} \rightarrow f$ a.e. $(\mu)$; (ii) $g_{n}, g \in L^{1}(\Omega, \mathcal{F}, \mu)$ and $\int\left|g_{n}\right| d \mu \rightarrow \int|g| d \mu$ as $n \rightarrow \infty$. Then, $f \in$ $L^{1}(\Omega, \mathcal{F}, \mu)$, $$ \lim _{n \rightarrow \infty} \int f_{n} d \mu=\int f d \mu \quad \text { and } \quad \lim _{n \rightarrow \infty} \int\left|f_{n}-f\right| d \mu=0 $$

Proof: By Fatou's lemma, $$ \int|f| d \mu \leq \liminf _{n \rightarrow \infty} \int\left|f_{n}\right| d \mu \leq \liminf _{n \rightarrow \infty} \int\left|g_{n}\right| d \mu=\int|g| d \mu<\infty $$ Hence, $f$ is integrable. For proving the second part, let $h_{n}=f_{n}+g_{n}$ and $\gamma_{n}=g_{n}-f_{n}, n \geq 1$. Then, $\left\{h_{n}\right\}_{n \geq 1}$ and $\left\{\gamma_{n}\right\}_{n \geq 1}$ are sequences of nonnegative integrable functions. By Fatou's lemma and (ii), $$ \begin{aligned} \int(f+g) d \mu &=\int \liminf _{n \rightarrow \infty} h_{n} d \mu \\ & \leq \liminf _{n \rightarrow \infty} \int h_{n} d \mu \\ &=\liminf _{n \rightarrow \infty}\left[\int g_{n} d \mu+\int f_{n} d \mu\right] \\ &=\int g d \mu+\liminf _{n \rightarrow \infty} \int f_{n} d \mu \end{aligned} $$ Similarly, $$ \int(g-f) d \mu \leq \int g d \mu-\limsup _{n \rightarrow \infty} \int f_{n} d \mu . $$ According to the linearity of integral,$\int(g \pm f) d \mu=\int g d \mu \pm \int f d \mu .$ Hence, $$ \int f d \mu \leq \liminf _{n \rightarrow \infty} \int f_{n} d \mu $$ and $$ \limsup _{n \rightarrow \infty} \int f_{n} d \mu \leq \int f d \mu $$ yielding that $\lim _{n \rightarrow \infty} \int f_{n} d \mu=\int f d \mu$. For the last part, apply the above argument to $f_{n}$ and $g_{n}$ replaced by $\tilde{f}_{n} \equiv\left|f-f_{n}\right|$ and $\tilde{g}_{n} \equiv g_{n}+|f|$, respectively.

This certificate is quoted from Measure Theory and Probability Theory. Krishna B. Athreya, Soumendra N. Lahiri.


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