# 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.