Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Royden (4th edition), it says one can prove the General Lebesgue Dominated Convergence Theorem by simply replacing $g-f_n$ and $g+f_n$ with $g_n-f_n$ and $g_n+f_n$. I proceeded to do this, but I feel like the proof is incorrect.

So here is the statement:

Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on $E$ that converge pointwise a.e. on $E$ to $f$. Suppose there is a sequence $\{g_n\}$ of integrable functions on $E$ that converge pointwise a.e. on $E$ to $g$ such that $|f_n| \leq g_n$ for all $n \in \mathbb{N}$. If $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $g_n$ = $\int_E$ $g$, then $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $f_n$ = $\int_E$ $f$.

$$\int_E (g-f) = \liminf \int_E g_n-f_n.$$

By the linearity of the integral:

$$\int_E g - \int_E f = \int_E g-f \leq \liminf \int_E g_n -f_n = \int_E g - \liminf \int_E f_n.$$


$$\limsup \int_E f_n \leq \int_E f.$$

Similarly for the other one.

Am I missing a step or is it really a simple case of replacing.

share|cite|improve this question
I guess that, instead of $\int_E (f-g) = \liminf \int_E g_n-f_n$ you mean $\int_E (f-g) = \liminf \int_E f_n-g_n$. – Leandro Oct 13 '11 at 1:24
up vote 17 down vote accepted

Since $|f_n| \leq g_n$ for all $n$ and $f_n$ ($g_n$ respectively) converge pointwise a.e. on $E$ to $f$ ($g$ respectively), we have $|f|\leq g$ pointwise a.e. on $E$. Therefore, for all $n$ we have $$|f_n-f|\leq g_n+g$$ pointwise a.e. on $E$. Now apply Fatou Lemma to the nonegative function $g_n+g-|f_n-f|$, we have $$\liminf_{n\rightarrow\infty}\int_E(g_n+g-|f_n-f|)\geq\int_E\liminf_{n\rightarrow\infty}(g_n+g-|f_n-f|).$$ The right hand side is equal to $$\int_E\liminf_{n\rightarrow\infty}(g_n+g-|f_n-f|)=2\int_Eg,$$ since $f_n$ ($g_n$ respectively) converge pointwise a.e. on $E$ to $f$ ($g$ respectively). On the other hand, the left hand side is equal to $$\liminf_{n\rightarrow\infty}\int_E(g_n+g-|f_n-f|)=2\int_Eg-\limsup_{n\rightarrow\infty}\int_E|f_n-f|$$ since $\displaystyle\lim_{n \rightarrow \infty}\int_Eg_n=\int_Eg$ by assumption. Now putting all these together, we obtain $$0\geq\limsup_{n\rightarrow\infty}\int_E|f_n-f|.$$ Since $\displaystyle\int_E|f_n-f|\geq\Big|\int_Ef_n-f\Big|$, by the above inequality we have $$0\geq\limsup_{n\rightarrow\infty}\Big|\int_E(f_n-f)\Big|\geq\liminf_{n\rightarrow\infty}\Big|\int_Ef_n-f\Big|\geq 0.$$ By the above equality, $\displaystyle\limsup_{n\rightarrow\infty}\Big|\int_E(f_n-f)\Big|=\liminf_{n\rightarrow\infty}\Big|\int_E(f_n-f)\Big|$, i.e. $\displaystyle\lim_{n\rightarrow\infty}\Big|\int_E(f_n-f)\Big|$ exists. Moreover, by the above equality again, $\displaystyle\lim_{n\rightarrow\infty}\Big|\int_E(f_n-f)\Big|=0$, which implies $$\lim_{n\rightarrow\infty}\int_Ef_n=\int_Ef,$$ as required.

share|cite|improve this answer
How does $\liminf_{n} \int (g_n+g-|f_n-f|) = 2 \int g- \limsup_n \int |f_n-f|$ ? We don't know that liminf is linear right ? – pikachuchameleon Oct 8 '15 at 22:21
@AshokVardhan: You are in right ,$\liminf $ is not linear in general. But if one of the sequences is convergence then it's linear. – user217174 Nov 4 '15 at 14:01

You made a mistake: $$\liminf \int (g_n-f_n) = \int g-\limsup \int f_n$$ not $$\liminf \int (g_n-f_n) = \int g-\liminf \int f_n.$$

Here is the proof:

$$\int (g-f)\leq \liminf \int (g_n-f_n)=\int g -\limsup \int f_n$$

which means that

$$\limsup \int f_n\leq \int f$$


$$\int (g+f)\leq \liminf \int(g_n+f_n)=\int g + \liminf \int f_n$$

which means that

$$\int f\leq \liminf \int f_n$$


$$\limsup \int f_n\leq \int f\leq \liminf\int f_n\leq \limsup \int f_n$$

So they are all equal.

share|cite|improve this answer
Your answer will be much easier to read if you write it using MathJax. Here is a tutorial:… – coldnumber Aug 1 '15 at 0:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.