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The question is

Let $(X, \cal M, \mu)$ be a finite measure space and fix $E\in \cal M$. For each $n \in \Bbb{N}$ define the function $f_n:(X, \cal M) \to \Bbb{R}$ given by $$ {f_n}(x) = \left\{ {\begin{array}{*{20}{c}} {{\chi _E}(x)}&{n{\rm{ \space is \space odd}}}\\ {1 - {\chi _E}(x)}&{n{\rm{ \space is \space even}}} \end{array}} \right. $$ for each $x\in X$.

Compute $\int_X {\liminf\limits_{n \to \infty } } {f_n}\,d\mu $ and $\liminf\limits_{n \to \infty } \int_X {{f_n}\,d\mu }$.

I don't know if I understand the question correctly, but it looks that $\liminf\limits_{n \to \infty } {f_n} = 0$?

So $\int_X \liminf\limits_{n \to \infty} {f_n}\,d\mu = 0$ ? And $\liminf\limits_{n \to \infty } \int_X {{f_n}\,d\mu } = \min \{ \mu (E),\mu (X\setminus E)\} $?

I strongly doubt my judgment but I could not see where I go wrong. Thank you!

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    $\begingroup$ Your answer is correct. $\endgroup$ – Aaron Aug 20 '15 at 3:25
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Your answer seems correct.

By $\liminf\limits_{n \to \infty } {f_n} = 0$ you easily deduce that $\int_X \liminf\limits_{n \to \infty} {f_n}\,d\mu = 0$, as you said.

And, if $n_k$ denotes the subsequence for odd $n$'s , we have $$ \int_X {f_{n_k}}\,d\mu=\mu(E) $$ for every $k$ and similarly, denoting by $n_j$ the subsequence of even indices, $$ \int_X {f_{n_k}}\,d\mu=\mu(X\setminus E), $$ for every $j$. From this you get your answer $\liminf\limits_{n \to \infty } \int_X {{f_n}\,d\mu } = \min \{ \mu (E),\mu (X\setminus E)\} $.

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