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Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous function.

[Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable? If not, can one find a counter-example?

Note that, for any $c$, we have $$(x: g(x) < c) = \text{Proj}_x ((x,u): g(x,uf(x,u) < c),$$ where $\text{Proj}_x$ is a projection operator to $x$-axis. In the context of measurable selection theorem, the projection of Borel set $((x,u): g(x,uf(x,u) < c)$ of $\mathbb{R}^2$ is not necessarily a Borel set of $\mathbb{R}$. But, I can not find a counter-example.

If there exists a proper counter-example, then it also implies that a semicontinuous real function is not necessarily Borel measurable.

Thanks.
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Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous function.

[Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable? If not, can one find a counter-example?

Note that, for any $c$, we have $$(x: g(x) < c) = \text{Proj_x} text{Proj}_x ((x,u): g(x,u) < c),$$ where $\text{Proj}_x$ is a projection operator to $x$-axis. In the context of measurable selection theorem, the projection of Borel set $((x,u): g(x,u) < c)$ of $\mathbb{R}^2$ is not necessarily a Borel set of $\mathbb{R}$. But, I can not find a counter-example.

If there exists a proper counter-example, then it also implies that a semicontinuous real function is not necessarily Borel measurable.

Thanks.
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Is a semicontinuous real function Borel measurable?

Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous function.

[Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable? If not, can one find a counter-example?

Note that, for any $c$, we have $$(x: g(x) < c) = \text{Proj_x} ((x,u): g(x,u) < c),$$ where $\text{Proj}_x$ is a projection operator to $x$-axis. In the context of measurable selection theorem, the projection of Borel set $((x,u): g(x,u) < c)$ of $\mathbb{R}^2$ is not necessarily a Borel set of $\mathbb{R}$. But, I can not find a counter-example.

If there exists a proper counter-example, then it also implies that a semicontinuous real function is not necessarily Borel measurable.

Thanks.