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Let ${f_{n}}$ be a sequence in $L^{2}(X,\mu)$ such that $||f_{n}||_{2} \rightarrow 0$ as $n \rightarrow \infty$. How to show that:

$\displaystyle \lim_{n \to \infty} \int_{X} |f_n(x)| \log(1+|f_{n}(x)|) d\mu = 0$

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Use the fact that $\log (1+y) \leq y$.

So we get $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \displaystyle \int_{X} |f_n(x)| |f_n(x)| d \mu = \displaystyle \int_{X} |f_n(x)|^2 d \mu = ||f_n||_2^2$.

Also, note that $\displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \geq 0$.

Hence, $0 \leq \displaystyle \lim_{n \rightarrow \infty} \displaystyle \int_{X} |f_n(x)| \log(1+|f_n(x)|) d \mu \leq \lim_{n \rightarrow \infty} ||f_n||_2^2 = 0$

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    But in order to carry integration on the LHS we need first to show that the LHS is integrable but it is because it is measurable and it less or equal than a function which is integrable (by assumption), so by a standard result the LHS is integrable, right?2010-11-19
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    @user10: Yes. If you have a measurable function $f$ and another measurable function $g$, then $g(f)$ is also measurable. Hence LHS is measurable and then you can argue as I said.2010-11-19