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I found this on a qualifier exam, and I think it will help me understand $L^p$ spaces better.

Let $f_n$ be a sequence of measurable function on a finite measure space. Suppose that $\sup_n \int_X |f_n(x)|^2 d\mu < \infty$ and that $\lim_{n\to \infty}f_n(x) =: f(x)$ exists $\mu$-almost everywhere. Which of the following are true (proving or providing a counterexample):

(1) $\int_X |f(x)|^2 d\mu < \infty$

(2) $ \int_X |f(x)| d\mu < \infty$

(3) $\lim_{n\to\infty} \int_X |f_n(x) - f(x)|^2 d\mu = 0$

(4) $\lim_{n\to\infty} \int_X |f_n(x) - f(x)| d\mu = 0$

1 Answers 1

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  1. It's true by an application of Fatou's lemma to $\{|f_n|^2\}$, a sequence of non-negative measurable functions.

  2. It's true by Cauchy-Bunyakovsky-Schwarz inequality (and the fact that the measure space is finite).

  3. Take $f_n(x):=\sqrt n \chi_{(0,n^{—1})}(x)$ on $[0,1]$; then $\int_{[0,1]}|f_n|^2dx=1$, and $f_n\to 0$ almost everywhere.

  4. It's true: apply Egoroff theorem and Cauchy-Bunyakovsky-Schwarz inequality. We can assume that $f=0$ (otherwise we consider $g_n:=f_n-f$ instead of $f_n$, which is integrable and $g_n\to 0$ pointwise). Fix $\varepsilon>0$. Then we can find $C$ measurable such that $\sup_{x\in C}|f_n(x)|\to 0$ and $\mu(X\setminus C)\lt \varepsilon$. We have $\int_X|f_n(x)|d\mu(x)\leqslant \mu(X)\sup_{x\in C}|f_n(x)|+\sqrt{\varepsilon}\sqrt{\int_X |f_n|^2}d\mu,$ so $\limsup_{n\to+\infty}\int_X|f_n(x)|d\mu(x)\leqslant \sqrt \varepsilon\sup_{k\in\Bbb N}\sqrt{\int_X |f_k|^2}d\mu.$

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    It's not so obvious: we use (1).2014-05-02