# Prove $\liminf\limits_{n\to\infty} F(x_n)\le F(x)$

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 $\int{f} \leq \liminf{\int f_n}$ – Euler....IS_ALIVE Aug 22 '12 at 20:16

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 need some more help. I see the first part is by right continuity, and since $F$ is nondecreasing $F(x_n) \leq F(y)$ for all n large enough from the second part. But I don't see how this shows $\lim \inf_{n \rightarrow \infty} F(x_n) = \sup \{ \inf {F(x_m) : m \geq n} : n \geq 1 \} \leq F(x)$. I'm I using the right definition of $\lim \inf$ or is there a theorem that will make my work easier? – mark Aug 22 '12 at 21:32
See Edit.   – Did Aug 22 '12 at 22:08
I think you meant to write $\lim \inf$. Thanks for the hint, you pretty much spelt it out for me! – mark Aug 23 '12 at 11:57
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.) – Did Aug 23 '12 at 12:12

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, $\liminf_k x_k=\min\{\liminf\sigma_L,\liminf\sigma_R\}$.

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