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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): $$\int fd\lambda=\lim_{n\to\infty}\int f_nd\lambda.$$

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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 at 11:25
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up vote 2 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.

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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
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