Is the integral of a measurable function measurable (wrt. a parameter)

Question

Let $u \in L^2(0,T;L^2(\Omega))$ on a compact Riemann manifold $\Omega$. Is it true that $$t \mapsto \int_\Omega u(t)w$$ is measurable for $w \in L^2(\Omega)$?

Answer 1

The map

$$ \Phi : L^2(\Omega) \to \Bbb{R}, f \mapsto \int_\Omega f \cdot w $$

is linear and continuous by the Cauchy-Schwarz inequality.

The map which you want to be measurable is given by

$$ \Phi \circ f : (0,T) \to \Bbb{R}, t \mapsto \Phi(f(t)) = \int_\Omega f(t) w. $$

Now $f : (0,T) \to L^2(\Omega)$ is measurable, so that $\Phi \circ f$ is measurable as the composition of a (continuous, hence) measurable function and a measurable function.