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I am reading Folland's, Real Analysis and I am stuck at the following exercise (2.33).

If $f_n \geq 0$ and $f_n \to f$ in measure, then $\int f \leq \liminf \int f_n$

It smells like Fatou's, however I couldn't see anything. Thanks in advance!

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1 Answer 1

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By the definition of $\liminf$, there is a subsequence $f_{n_k}$ of $f_n$ such that: $$ \lim_{k \to \infty} \int f_{n_k} = \liminf_{n \to \infty} \int f_n $$

$f_{n_k}$ also converges in measure to $f$. To see why, notice that the sequence $\mu\left(\left\{ x: \left| f_n(x) - f(x) \right|\geq \epsilon \right\}\right)$ converges to $0$ as $n \to \infty$. Thus, its subsequence $\mu\left(\left\{ x: \left| f_{n_k}(x) - f(x) \right|\geq \epsilon \right\}\right)$ also converges to $0$ as $k \to \infty$. This is a basic result in sequence convergence.

By theorem 2.30 in Folland's book, there is a subsequence of $f_{n_k}$ that converges pointwise a.e. to $f$. Call it $f_{n_{k_j}}$. Notice that it also satisfies: $$ \lim_{j \to \infty} \int f_{n_{k_j}} = \liminf_{n \to \infty} \int f_n $$

Apply Fatou's lemma to $\{f_{n_{k_j}}\}$, and put it all together to find that: $$ \int f = \int \lim_{j \to \infty} f_{n_{k_j}} \le \lim_{j \to \infty} \int f_{n_{k_j}} = \liminf_{n \to \infty} \int f_n $$

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Thanks for helps. But I have one question. How can we verify that if $f_n \to f$ in measure, then $\exists$ a convergent subsequence (if not itself) $f_{n_k} \to f$ also? Is there a specific theorem for this (or is it obvious)? –  Deniz Jan 13 '13 at 14:50
    
In particular, rigorously, we define the convergence in measure, $$\mu\left( \left\lbrace x: \left\lvert f_n(x) - f(x) \right\rvert \geq \epsilon \right\rbrace \right) \to 0$$as $n \to \infty$. Then, how can we say this for $f_{n_k}$ and we say this for as $n \to \infty$ or $k \to \infty$ or both? –  Deniz Jan 13 '13 at 14:56
    
@John This is theorem 2.30 in Folland's book. Convergence in measure implies the existence of a subsequence that converges pointwise a.e.. –  Ayman Hourieh Jan 13 '13 at 18:29
    
OK thanks, then I guess right after the equation you have a typo? You wrote: "Since $f_{n_k}$ also converges in measure to $f$...". I ask do we know this subsequence also converges in measure? Thm. 2.30 says the subsequence converges a.e. to $f$, not in measure as far as I remember. –  Deniz Jan 13 '13 at 18:40
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@John Happy to help! –  Ayman Hourieh Jan 13 '13 at 18:58

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