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Let $(X,\mathcal M,\mu)$ be a measure space such that $ \mu(X)<\infty$. Suppose that $(f_n)$ is a sequence in $L^p(X)$ such that $|f_k|$ converges weakly to $|f|$ in $L^p(X)$. A solution to a problem I found in a book (Problems and Solutions in Mathematics - Major American Universities PHD Qualifying Questions and Solutions) says that it follows by the Vitali-Hahn-Saks for any $ \varepsilon >0$ there exists $ \delta>0$ such that $$ \int_E |f_k| d\mu + \int_E |f| d \mu < \varepsilon$$ whenever $\mu(E)<\delta$.

My question is:

What is the Vitali Hahn Saks Theorem (I remember seeing it, but I forgot where, and Google didn't help much) and how does the theorem apply here.

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I don't understand why they use this theorem. Since $|f_k|$ converge weakly to $|f|$ and $\mathbf 1_X$ is integrable, we have for a fixed $\varepsilon$ an integer $n_0$ such that if $k\geq n_0$ then $\int_X |f_k|d\mu\leq\int_X |f|d\mu +\frac{\varepsilon}2$. We can find a $\delta>0$ such that $\int_E |f|d\mu\leq \frac{\varepsilon}2$ if $\mu(E)\leq \delta$ (first approximate $|f|$ by a simple function), so we have $\int_E(|f_k|+|f|)d\mu\leq \varepsilon$ for $k\geq n_0$. Apply what we did for $|f|$ to $|f_k|$ for $k<n_0$ to get the wanted inequality. –  Davide Giraudo Jan 3 '12 at 20:38
    
@DavideGiraudo: I figured there is an easier way, but I was curious how that theorem applies. –  Beni Bogosel Jan 3 '12 at 23:01

1 Answer 1

up vote 3 down vote accepted

The version of Vitali-Hahn-Saks they are using is found e.g. in Yosida, "Functional Analysis", sec. II.2. The measures $|f_k| \ d\mu$ and $|f| \ d\mu$ are absolutely continuous wrt $\mu$, and the theorem asserts that this absolute continuity is uniform in $\mu$.

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Thank you for your quick answer. –  Beni Bogosel Jan 3 '12 at 23:01

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