Dominated convergence theorem and bounded functions

Question

Consider the following setting for the application of the bounded convergence theorem (BCT). Let $f_n$ be a sequence of measurable $\mu$-integrable functions such that $f_n\rightarrow f$ pointwise. I understand that BCT says that if $\sup_n \sup_xf_n\leq M$ for some finite constant $M$ then $\int |f_n-f|d\mu \rightarrow 0$.

In Jesse Madnick's comment below the question Explanation of the Bounded Convergence Theorem it seems to be suggested that it is possible to relax the condition $\sup_n \sup_xf_n\leq M$ also to the setting where all the $f_n$ are bounded functions (but the bound is not uniform across them?). Is my understanding correct - and do you have any proof or reference for this? Many thanks!

A very important hypothesis in that link is the finiteness of $\mu$. Without this even uniform boundedness is not enough. — Kavi Rama Murthy, Mar 11 '23 at 12:20

score 2 · Accepted Answer · answered Mar 11 '23 at 12:22

2

The example $f_n=n\chi_{(0,\frac 1n )}, f=0$ on $(0,1)$ with Lebesgue measure shows that boundedness of each $f_n$ is not enough.

Jessee Madnick's commnent in the link assumes uniform convergence which you don't have.

answered Mar 11 '23 at 12:22

Kavi Rama Murthy

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Dominated convergence theorem and bounded functions

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