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Suppose $F$ is a nondecreasing and right continuous function, and the sequence $\{x_n\}_{n\geq1}$ converges to $x$. Then $\liminf\limits_{n\to\infty}F(x_n)\leq F(x)$.

How can I prove this?

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    It is interesting to contrast this with Fatou's lemma, which says that under certain conditions (including $f_n \rightarrow f$ a.e), you get $\int{f} \leq \liminf{\int f_n}$2012-08-22

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Hints:

  • For every $z\gt F(x)$, there exists $y\gt x$ such that $F(u)\leqslant z$ for every $u\leqslant y$.
  • Since $x_n\to x$, $x_n\leqslant y$ for every $n$ large enough.
  • Hence...

Edit:

... $F(x_n)\leqslant F(y)\leqslant z$ for every $n$ large enough. In particular, $\limsup\limits_{n\to\infty}F(x_n)\leqslant z$. This is valid for every $z\gt F(x)$, hence...

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    I meant limsup. (By the way, the result with liminf is a little odd because, if it holds, one can show that the result with limsup also holds.)2012-08-23
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HINT: Let $\sigma_R=\langle x_{n_k}:k\in\Bbb N\rangle$ be the subsequence of terms greater than or equal to $x$, and let $\sigma_L=\langle x_{m_k}:k\in\Bbb N\rangle$ be the subsequence of terms less than $x$.

  1. If either subsequence is finite, it can be ignored.

  2. If $\sigma_R$ is infinite, $\lim\limits_{k\to\infty}F(x_{n_k})=F(x)$.

  3. For each $k\in\Bbb N$, $F(x_{m_k})\le F(x)$.

  4. If both subsequences are infinite, $\liminf_k x_k=\min\{\liminf\sigma_L,\liminf\sigma_R\}$.