# General Lebesgue Dominated Convergence Theorem

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 $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $f_n$ = $\int_E$ $f$.

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

By the linearity of the integral:

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

So,

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

Similarly for the other one.

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

• 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

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.

• 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
• Very nice. Using the same proof, one can prove the "stronger" result: $\lim\int_E |f_n-f|=0$. – yoyostein Sep 20 '16 at 7:51
• @yoyostein does that mean fn converges to f in L1. Also is there a similar statement for convergence in Lp ? – Namch96 Apr 8 '17 at 17:56
• @Namch96 yes and yes – yoyostein Apr 9 '17 at 1:07

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

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

Also

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

which means that

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

i.e.

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

So they are all equal.

I ignore the E and proceed to prove it as required by Exercise 2.20 in Folland.

The problem is essentially a Generalized Dominated Convergence Theorem which I prove by reworking the proof of Lebesgue's Dominated Convergence Theorem introduced in class.

$|f_n| \le g_n$ implies $-g_n \le f_n \le g_n$ which implies $f_n + g_n \ge 0$ and $g_n-f_n \ge 0$

We know $\lim (f_n + g_n) = f + g$ and $\lim (g_n - f_n) = g -f$. We can therefore apply Fatou's Lemma to the above cases to get:

(I) $\int g + \int f = \int (g+f) = \int \lim (f_n + g_n) = \int \liminf (f_n + g_n) \le \liminf \int (f_n + g_n) = \liminf \int g_n + \liminf \int f_n = \int g + \liminf \int f_n$

Thus, $\int f \le \liminf \int f_n$.

(II) $\int g - \int f = \int (g-f) = \int \lim (g_n - f_n) = \int \liminf (g_n - f_n) \le \liminf \int (g_n - f_n) = \liminf \int g_n + \liminf \int - f_n = \int g - \limsup \int f_n$

Thus, $\int f \ge \limsup \int f_n$.

(III) Notice $\{ \int f_n \}_{n \in \mathbb{N}}$ is fundamentally a sequence. So, as is true for any sequence: $$\liminf \int f_n \le \limsup \int f_n$$

(IV) Also, our work with Fatou's has established the following: $$\limsup \int f_n \le \int f \le \liminf \int f_n$$

Combining (III) and (IV), we get $\liminf \int f_n = \limsup \int f_n = \int f$ . The first equality implies the existence of $\lim \int f_n$ and the second establishes: $$\lim \int f_n = \int f = \int \lim f.$$

• this $$\liminf \int (f_n + g_n) = \liminf \int g_n + \liminf \int f_n$$ is not right – Masacroso Sep 30 '18 at 15:57