Prove that $x_n\to x\implies \liminf\limits_{n\to\infty}f(x_n)\leq f(x)$

Question

Let $\{f_n\}$ be a sequence of real-valued functions on $[0,1]$ and let $M$ be a real number such that:

(i) for each $n,$ $f_n$ is continuous; (ii) for each $x\in[0,1]$ and for each $n,$ $f_n(x)\leq f_{n+1}(x)\leq M.$

Prove that $\{f_n\}$ converges pointwise on $[0,1]$ to a function $f$ which is lower semicontinuous.

(that is for all $\{x_n\}\subset [0,1]$ and for all $x\in [0,1],$ $x_n\to x\implies \liminf\limits_{n\to\infty}f(x_n)\leq f(x).$

Please, can anyone help me out with this?

Answer 1

Since $\{f_n\}$ is increasing and bounded above $f(x)=lim_n f_n(x)$ exists. Suppose $\lim \inf f(x_n) >f(x)$. Choose $\epsilon >0 $ such that $\lim \inf f(x_n) >f(x)+\epsilon $. There exists $n_0$ such that $f(x_n) >f(x)+\epsilon $ for $n \geq n_0$. Hence $f_k (x_{n_0}) >f(x)+\epsilon $ for some $k$. But $f(x) \geq f_k (x_{n_0})$ and this gives a contradiction.