Measurability of composition (Bochner function)

Question

Let $g \in L^2(0,T;X)$ and $f:[0,T]\times X \to \mathbb{R}$ is such that $$\frac{d}{dt}F(t,x) = f(t,x).$$ so $t \mapsto f(t,x)$ is measurable on $[0,T]$ (but I don't know in what sense) for fixed $x$.

Now I believe that the composition $t \mapsto f(t,g(t))$ is measurable only if $f$ is Borel measurable and $g$ is Lebesgue or Borel measurable.

But I have no idea what sense $f$ and $g$ are measurable. According to wiki, $t \mapsto g(t)$ is Bochner measurable but I don't know any idea how this is related to Borel or Lebesgue.

Can anyone advise me? Thank you.

Answer 1

A sufficient condition for the measurability of $t \mapsto f(t,g(t))$ is the Caratheodory condition of $f$:

• $f(\cdot, x) : [0,T] \to \mathbb{R}$ is measurable for all $x \in X$
• $f(t, \cdot) : X \to \mathbb{R}$ is continuous for almost all $t \in [0,T]$

Since $g$ is Bochner-measurable, this implies the measurability of $t \mapsto f(t,g(t))$.

As a reference, have a look for Goldberg, Kampowski, Tröltzsch: On Nemytskii operators in Lp-spaces of abstract functions, Math. Nachrichten 155 (1992), 127-140.