# Necessity of generalization of Dominated Convergence theorem

In Royden's Real Analysis there's this generalization of Lebesgue's Dominated Convergence theorem (p.92):

Let $\{g_n\}$ be a sequence of integrable functions which converges a.e. to an integrable function g. Let $\{f_n\}$ be a sequence of measurable functions such that $|f_n| \leq g_n$ and $\{f_n\}$ converges to f a.e. If $$\int g = \lim \int g_n$$ then $$\int f = \lim \int f_n$$

Are there examples where the usual version, where there's just measurable g rather than the sequence $\{g_n\}$, isn't enough and it's necessary to use this generalization? I've never heard of any but it would be pretty interesting to find one.

## 2 Answers

Let's start out with a sequence of nonnegative integrable functions $g_n$ whose supremum is not integrable, but which converge a.e. to an integrable function $g$ and whose integrals converge to the integral of $g$. For example, on $[0,1]$ we might take $g_n(x) = 1 + J_n(x)/x$ where $J_n$ is the indicator function of the interval $[1/(n+1), 1/n]$, and $g(x) = 1$.

Take any measurable $f_n \to f$ with $|f_n| \le g_n$ and converging a.e. to some measurable function $f$.

Note: I'm not claiming that it's "necessary" to use the generalization, just that it's not a completely obvious application of the standard LDCT.

This is to elaborate on the comments of Robert Israel at the end of his answer.

If $$f_n\xrightarrow{n\rightarrow\infty}f$$ pointwise $$\mu$$-a.s., $$|f_n|\leq g_n$$, and $$\lim_n\int g_n\,d\mu=\int \lim g_n\,d\mu$$, then the "generalized" dominated convergence implies that $$\int|f-f_n|\,d\mu\xrightarrow{n\rightarrow\infty}0$$, for $$|f-f_n|\leq |f|+ |g_n|$$, and $$\lim_n\int(|f|+g_n)\,d\mu=\int\lim_n(|f|+g_n)\,d\mu=\int(|f|+\lim_n|g|)\,d\mu$$

Conversely, if $$\lim_n\int|f_n-f|\,d\mu=0$$, then \begin{align} \int \big||f| -|f_n|\big|\,d\mu&\leq\int |f - f_n|\,d\mu\xrightarrow{n\rightarrow\infty}0\\ \Big|\int (f-f_n)\,d\mu\Big|&\leq \int |f - f_n|\,d\mu\xrightarrow{n\rightarrow\infty}0 \end{align}

Thus, if $$f_n\rightarrow f$$ $$\mu$$-a.s., the existence of a sequence $$g_n\geq0$$ such that $$|f_n|\leq g_n$$ and $$\lim_n\int g_n\,d\mu=\int \lim_n g_n\,d\mu$$ is necessary and sufficient for $$\lim_n\int|f-f_n|\,d\mu=0$$, and all the nice consequence this entails (necessity follows by taking $$g_n=|f_n|$$, sufficiency is the generalized theorem).

I can't answer right now whenever the classical version of dominated convergence is also a necessary conditions, that is if $$f_n\xrightarrow{n\rightarrow\infty}f$$ $$\mu$$-a.s. and $$\lim_n\int|f-f_n|\,d\mu$$ implies that there is $$g\in L_1$$ such that $$|f_n|\leq g$$ for all $$n$$. This can happen along a subsequence $$f_{n_k}$$ though. Take for example a subsequence $$f_{n_k}$$ such that $$\int|f_{n_{k+1}}-f_{n_k}|\,d\mu\leq \frac{1}{k^2}$$ Define \begin{align} g_k&=f_{n_1}+\sum^k_{j=1}(f_{n_{j+1}}-f_{n_j})\\ G_k&=|f_{n_1}|+\sum^k_{j=1}|f_{n_{j+1}}-f_{n_j}| \end{align} By monotone convergence, $$G_k$$ converges to an integrable function $$G=\sup_kG_k$$ and $$\int|G-G_k|\,d\mu\xrightarrow{k\rightarrow\infty}0$$. Notice that $$g_k=f_{n_{k+1}}$$, and $$|g_k|\leq G_k\leq G$$.

Finally, if the convergence of $$\lim_n\int|f_n-f|\,d\mu=0$$ is fast enough, for example, there is a sequence $$a_n>0$$ such that $$\sum_na_n<\infty$$ and $$\int|f_n-f|\,d\mu\leq a_n$$, then the argument introduced above shows that an integrable dominating function $$G$$, i.e. $$|f_n|\leq G$$ for all $$n$$, does exists.