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?
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?
Hints:
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...
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 $x$.
If either subsequence is finite, it can be ignored.
If $\sigma_R$ is infinite, $\lim\limits_{k\to\infty}F(x_{n_k})=F(x)$.
For each $k\in\Bbb N$, $F(x_{m_k})\le F(x)$.
If both subsequences are infinite, $\liminf_k x_k=\min\{\liminf\sigma_L,\liminf\sigma_R\}$.