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Suppose $f_n$ is a sequence of non-negative measurable functions that converges in measure to a measurable function $f$. I am trying to show that

$$\int f\leq\liminf_n\int f_n$$

Using Fatou's lemma we know that

$$\int\liminf_n f_n\leq\liminf_n\int f_n$$

so it would be enough to show that $$f\leq\liminf_n f_n$$ but, is this even true? am I going in the right direction?

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  • $\begingroup$ Do you assume that $f_n$ is non-negative? $\endgroup$
    – saz
    Commented Jan 23, 2017 at 10:38
  • $\begingroup$ Erm yes, sorry, I'll edit. $\endgroup$
    – Smurf
    Commented Jan 23, 2017 at 10:49

3 Answers 3

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Hints:

  1. There exists a subsequence $(f_{n_k})_k$ of $(f_n)_n$ such that $$\liminf_{n \to \infty} \int f_n = \lim_{k \to \infty} \int f_{n_k}.$$
  2. Check that $f_{n_k} \to f$ in measure as $k \to \infty$.
  3. There exists a further subsequence, say $(g_j)_j$, of $(f_{n_k})_k$ such that $g_j \to f$ almost everywhere. Applying Fatou's lemma yields, $$\int f \leq \liminf_{j \to \infty} \int g_j= \liminf_{n \to \infty} \int f_n.$$
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  • $\begingroup$ Then I get something like $$\int f=\int \liminf_k f_{n_k}\leq \liminf_k\int f_{n_k}$$ which does not seem to lead to anything. $\endgroup$
    – Smurf
    Commented Jan 23, 2017 at 10:21
  • $\begingroup$ @Smurf See my edited answer. $\endgroup$
    – saz
    Commented Jan 23, 2017 at 13:38
  • $\begingroup$ I don't think 1. is true. Although you gave me the idea for the proof, I will write it here in a second. $\endgroup$
    – Smurf
    Commented Jan 23, 2017 at 15:11
  • $\begingroup$ @Smurf Why do you think that it is not true? For any sequence $(a_n)$ there exists a subsequence $a_{n_k}$ such that $$\liminf_n a_n = \lim_k a_{n_k}.$$ Use this for $a_n := \int f_n$. $\endgroup$
    – saz
    Commented Jan 23, 2017 at 15:13
  • $\begingroup$ Because $$\liminf_n a_n=\inf\{\lim a_{n_k}\mid \text{ for every subsequence }a_{n_k}\}$$ it may not be a minimum, so 1. wouldn't hold. $\endgroup$
    – Smurf
    Commented Jan 23, 2017 at 15:19
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Let $\varepsilon>0$, then there is a subsequence $f_{n_k}$ such that $$\liminf\int f_n+\varepsilon\geq \lim\int f_{n_k}$$ since $f_{n_k}\rightarrow f$ in measure there is a subsequence $f_{n_{k_j}}$ such that $f_{n_{k_j}}\rightarrow f$ a.e., thus, by Fatou's lemma $$\int f=\int\lim f_{n_{k_j}}\leq \lim\int f_{n_{k_j}}=\lim\int f_{n_k}\leq\liminf \int f_n+\varepsilon$$

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Consider $g_n(x)=\inf\limits_{k \geq n}f_k(x)$, if so, we obtain $g_1 \leq g_2 \leq \ldots \leq g_n \leq \ldots$ and $\lim\inf f_n(x) \leq \lim\limits_{n \to \infty} g_n(x)$. By monotone convergence $\int\limits_X f(x)dx = \lim\limits_{n \to \infty} \int\limits_X g_n(x)dx$, which gives $g_n(x) \leq f_n(x)$, so $\int\limits_X g_n(x) dx \leq \int\limits_X f_n(x)dx$.

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  • $\begingroup$ $\liminf f_n=\lim g_n$ by definition. When you use monotone convergence theorem you are assuming that $g_n\rightarrow f$ a.e., which is not the case. $\endgroup$
    – Smurf
    Commented Jan 23, 2017 at 11:57

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