Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to prove that LDCT(Lebesgue Dominated Convergence Theorem) continues to hold if I replace the hypothesis $f_n \to f$ (convergence pointwise) with $f_n\to f$ (convergence in measure): $$\int fd\lambda=\lim_{n\to\infty}\int f_nd\lambda.$$

share|cite|improve this question
This is not exactly an extension, since convergence in measure does not imply convergence almost everywhere (only that there is an almost everywhere convergent subsequence). – tomasz Mar 9 '14 at 11:25

1 Answer 1

up vote 3 down vote accepted

Call $(X,\cal F,\mu)$ the involved measure space. Let $g$ integrable such that $|f_n(x)|\leqslant g(x)$ for almost every $x$.

As $g$ is integrable, denote $X':=\{g\neq 0\}=\bigcup_{n\geqslant 1}\{x,|g(x)|>n^{-1}\}$. Then $X'$ with the induced measure is $\sigma$-finite. Applying this version of dominated convergence theorem, we get that $$\int_{X'}fd\mu=\lim_{n\to +\infty}\int_{X'}f_nd\mu.$$ As $X\setminus X'=\{g=0\}\subset \{f=0\}\cup\bigcap_{n\geqslant 1}\{f_n=0\}$, we have $\int_{X'}fd\mu=\int_Xfd\mu$.

So fore each $n$, $\int_{X\setminus X'}fd\mu=\int_{X\setminus X'}f_nd\mu=0$, giving the wanted result.

share|cite|improve this answer
Is it true on $R^n$? – 89085731 Nov 17 '12 at 22:45
Yes (and in an arbitrary measured space when we have a non-negative measure). – Davide Giraudo Nov 17 '12 at 22:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.