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Let $\{f_{n}\}_{n=1}^{\infty}$ be a sequence of continuous functions on $\mathbb R$ , $f_n\geq 0$ for all $n$, where $|f_{n}(x)|\leq A$ for all $n=1,2,3,...$ on $\mathbb R$. Assume that $\lim_{n\to \infty}f_{n}(x)=0$ uniformly on $\mathbb R$, and $\int_{-\infty}^{\infty}|f_{n}(x)|dx<\infty$ for all $n=1,2,3,...$ . Is it true that $$ \lim_{n\to \infty}\int_{-\infty}^{\infty}|f_{n}(x)|dx=\int_{-\infty}^{\infty}\lim_{n\to \infty}|f_{n}(x)|dx=0 $$

The Dominated convergence, Monotone, and Fatou's theorems doesn't apply here. Any help?

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2 Answers 2

This is not true. The functions $f_n$ defined by $f_n(x)=\max\lbrace 0,\frac1n-\frac1{n^2}|x|\rbrace$ are a counter-example, as simple calculations should show. (The integral equals $1$ for each of these functions.)

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As Dejan's example shows, the problem is that because $\mathbb R$ has infinite measure, the functions $f_n$ can tend to $0$ uniformly, but still retaining their mass (hence becoming fatter and fatter).

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So I think it is the same in case of infinite summation $\sum_{n=1}^{\infty}$ instead of integration! –  Ben Jun 9 '12 at 2:05
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