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I have some problems about solving an exercise:

Prove that one function $f \colon [a,b] \to \mathbb{R}$ is lower semi-continuous if and only if, for all $x \in [a,b]$, we have $f(x)=\sup \{g(x) \mid g \in C[a,b] \text{ and } g \le f \text{ over } [a,b] \}\;.$

Assuming true that formula, I had no problems showing that $f^{-1}((t,\infty \ ])$ is open, using the property of supremum and the continuity of $g$'s.

I have difficulties proving the opposite implication. Beacuse $f \ge g$, we have that $f(x) \ge \sup{g(x)}$, but I am not able to show the other inequality.

Thank you.

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This would be an indirect solution, but I think what you have now implies that $\limsup_{x\to x_{0}} f(x)\le f(x_0)$ which is another equivalent definition of upper semi-continuity, and then use that definition to prove the implication that $f^{−1}((t,∞ ])$ is open.

In other words you may not even need to prove the reverse inequality.