Let $X$ be a compact pseudo-metric space and $f:X\to \mathbb R$ be lower semi-continuous. Then is $f(X)$ a Borel set?
The continuous image of a Borel set need not be Borel (there are projections of Borel sets in $\mathbb R^2$ which are not Borel in $\mathbb R$), and injective images of Borel sets under Borel maps are Borel. Here, the map isn't injective, but we are taking the image of the entire domain (and the domain is compact pseudo-metric). If $f$ were to be continuous then the result is true because the continuous image a compact set is compact, and hence closed in $\mathbb R$.
If $X$ is merely a compact topological space, then the result is false. Indeed, for arbitrary $f:X\to\mathbb R$, the coarsest topology that makes $f$ lower semi-continuous ($\{f^{-1}(a,\infty):a\in\mathbb R\}$) is compact when $f$ attains its minimum (the infimum is an indiscrete point). The fact that $X$ is pseudo-metric is important.
Related: Is the image of a $G_\delta$ set under a continuous mapping of $\mathbb R^n$ a Borel set?, Continuous image of $\mathbb{R}$ is Borel.
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