# Dominated Convergence Theorem for $p=\infty$?

Is there some version of DCT for $p=\infty$. That is, is it true that if there is a sequence of measurable functions defined on an open set $\Omega$ in $\mathbb{R}^n$, $f_n$ converging pointwise to a function $f$ such that there exists $g\in L^{\infty}(\Omega)$ with $|f_n(x)|\leq|g(x)|$, then $f_n,f\in L^{\infty}(\Omega)$ and $||f_n-f||_{\infty}\to 0$?

If this is indeed true, could someone point out the proof?

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It's not true. Consider the sequence $(f_n)$ defined by $f_n=\chi_{(0,1/n)}$ in $L^\infty(0,1)$. – David Mitra Sep 11 '12 at 21:19

In $\Bbb R$, this version doesn't work: take $f_n:=\chi_{(n,n+1)}$. We have $|f_n|\leq 1=:g\in L^{\infty}$ and $\lVert f_n-f\rVert_{\infty}=1$ for all $n$, with $f=0$.

Actually, the aim of dominated convergence is to switch a limit with an integral. When we work with $\lVert\cdot\rVert_{\infty}$, we don't use integrals (only properties of the measure).

The domination, for a sequence in $L^{\infty}$, means that the sequence of the norms is bounded, and pointwise convergence doesn't improve the things.

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do you mean $f=0$? – user38404 Sep 11 '12 at 21:31
@Vivek Yes, it was a typo. Thanks! – Davide Giraudo Sep 11 '12 at 21:36

No, it's false.

Take $$f_n = \begin{cases} 1 : & n < x < n+1 \\ 0 : &\textrm{otherwise}\end{cases}$$

The sequence $f_n$ converges pointwise to $0$, and is dominated by $g(x) \equiv 1 \in L^\infty$, but $||f_n - f||_\infty \equiv 1$ (so does not go to $0$).

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