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 2


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.


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