Is $\omega\mapsto\int f\;d\delta_\omega$ measurable?

Question

Let $(\Omega,\mathcal A)$ be a measurable space, $f:\Omega\to[0,\infty]$ be $\mathcal A$-measurable. Is $$\kappa:\Omega\to[0,\infty]\;,\;\;\;\omega\mapsto\int f\;d\delta_\omega$$ $\mathcal A$-measurable? ($\delta_\omega$ denotes the Dirac measure on $(\Omega,\mathcal A)$)

I've no idea how I could start. I suppose it's an easy conclusion from one or two basic results.

Answer 1

To elaborate on @Daniel Fischer's comment, for $A\in\mathcal A$, $$\delta_\omega(A)=\begin{cases}1,& \omega\in A\\ 0,& \omega\notin A,\end{cases}$$ it follows that $$\kappa(\omega) = \int_\Omega f\ \mathsf d\delta_\omega = f(\omega), $$ since trivially $\omega\in\Omega$. As $f$ was assumed to be $\mathcal A$-measurable, so too is $\kappa$.