Why can't I use dominated convergence theorem .

Question

$\{f_n(x)\}_{n\geq1}$ and $\{g_n(x)\}_{n\geq1}$ are both Lebesgue integrable functions and $|f_n(x)|\leq g_n(x)$ , if $f_n(x)$ convergence to $f(x)$ a.e. in $E$, $g_n(x)$ convergence to $g(x)$ a.e. in $E$ , and if $\lim\limits_{n\to \infty}\int_E g_n(x)dx=\int_E g(x)dx$ I need to prove $\lim\limits_{n \to \infty}\int_E f_n(x)dx=\int_E f(x)dx$

Why can't I use dominated convergence theorem like this. $|f(x)|\leq g(x)$ and $g(x)$ is integrable ,so $\lim\limits_{n\to \infty}\int_E f_n(x)dx=\int_E f(x)dx$

Answer 1

Your question has been answered in the comment but I suppose you want a valid proof of the original question. You need the additinal hypothesis that $\int g <\infty $. First note that if the result holds when $f_n \geq 0$ for all $n$ the we can apply the result for $f_n^{+}$ and $f_n^{-}$ to complete the proof. So let us assume that $f_n \geq 0$. Now $\int f(x) dx \leq \liminf \int f_n(x) dx$ by Fatou's Lemma. Now, $g_n-f_n \geq 0$ so $\int (g-f) \leq \liminf \int (g_n-f_n)=\int g - \limsup \int f_n$. This gives $\limsup \int f_n \leq \int f$. Combined with $\int f(x) dx \leq \liminf \int f_n(x) dx$ this completes the proof.

Answer 2

This theorem is very similar to DCT; infact it implies the DCT.

To use the dominated convergence theorem, we need to find a $g$ such that \begin{align} |f_n(x)| \leq g(x) \text{ a.e. }\end{align} for every $n \in \mathbb{N}$. It DOES NOT suffice to have \begin{align} |f(x)| \leq g(x) \end{align} Indeed take $f_n(x) = n \chi_{[0,1/n]}$ then $f_n(x) \rightarrow 0$, but $\int f_n(x) = 1$ (with respect to the lebesgue measure). But we have \begin{align} |f(x)| \leq \chi_{[0,1]}(x) \text{ a.e } \end{align} So we see your stated condition is not enough for DCT.

Now to prove the generalized dominated convergence theorem, we need to use Fatou's lemma. Indeed, define $h_n(x) = g_n(x) - |f_n(x)| \geq 0$.Then we see we can apply fatou's lemma, so we have \begin{align} \int h \text{ d}\mu \leq \liminf \int h_n \text{ d}\mu = \liminf \int g_n - |f_n| = \liminf \int g_n - \limsup \int |f_n| \text{ d}\mu\end{align} Then as $g_n \rightarrow g$ in $L^1$, we see that $\liminf \int g_n = \int g$. Then as $h = g - |f|$, we see that \begin{align} \int g - |f| \leq \int g - \limsup\int |f_n| \Rightarrow \int|f| \geq \limsup \int|f_n| \end{align} But by Fatou's lemma as $|f| \geq 0$, we have \begin{align} \liminf \int |f_n| \geq \int|f| \geq \limsup \int|f_n| \end{align} Hence, we see that $\int |f| = \lim_{n \rightarrow \infty} \int |f_n|$ as desired.