Proof of $\Phi^\alpha =\{x\in\mathbb R\mid f(x)\leq \alpha \}$ bounded imply $f$ lower semi continuous.

Question

Let $f:\mathbb R\to \mathbb R$. We suppose that for all $\alpha \in\mathbb R$, $\Phi^\alpha =\{x\in\mathbb R\mid f(x)\leq \alpha \}$ closed. Prove that $f$ is Lower semi continuous. I think that my proof is very complicated, and I was wondering if there is easier or not ?

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Proof Let $(x_n)$ a sequence that converge to $x_0$. Suppose by contradiction that $\liminf_{n\to \infty }f(x_n)<f(x_0)$. Set $y_n=\inf_{k\geq n}x_k$ and $\ell=\liminf_{n\to \infty }f(x_n)$. Let $\alpha \in\mathbb R$ s.t. $$\ell<\alpha <f(x_0).$$ By definition of $y_n$, for all $n\in\mathbb N^*$ there is $k_n\geq n$ s.t. $$f(x_{k_n})\leq y_n+\frac{1}{n}\leq \ell+\frac{1}{n},$$ where the last inequality comes from the fact that $(y_n)$ is increasing. So, let $N$ s.t. $\frac{1}{n}\leq \alpha -\ell$ for all $n\geq N$. In particular, $$f(x_{k_n})\leq \alpha <f(x_0),$$ for all $n\geq N$, and thus $(x_{k_n})_{n\geq N}$ is a sequence of $\Phi^\alpha $ that doesn't converges in $\Phi^\alpha $. Contradiction.

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I think my proof is quite complicate. Is there easier ?

Answer 1

A very little simplification could be to remark that $$y_n+\frac{1}{n}\underset{n\to \infty }{\longrightarrow } \ell<\alpha,$$ and thus, there is $N$ s.t. $$f(x_{k_n})\leq y_n+\frac{1}{n}<\alpha $$ for all $n\geq N$. So no need to use the fact that $(y_n)$ is increasing. But except this point, I'm not sure there is an easier proof. Well, I don't think that it's a very complicate proof neither...