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

Question

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?

Answer 1

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...

Answer 2

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