expectation of random probability measure

Question

Given $mu(omega, B)$ is a random probability measure on $mathbb{R}_{geq 0}$. This means for each $B$, it is a random variable, and for each $omega$ it is a probability measure.
How could I show that
$$mathbb{E}bigg[int_mathbb{R} f dmubigg] =int_mathbb{R} f dmathbb{E}[mu]$$
where $f$ is a bounded continuous function?
If $f$ is also differentiable, using Fubini theorem, we can rewrite
$$int_mathbb{R} f dmu = int_0^infty f dmu=int_0^infty f'(x)mu(x,infty) dx.$$
Is there any measurability issue for $mu(x,infty)$ when switching the expectation with the integral in the term below
$$mathbb{E}bigg[int_0^infty f'(x)mu(x,infty) dxbigg]?$$

Kavi Rama Murthy · Answer

Measurability is not an issue but your second equality needs justification.
There is a  much simpler way. The required identity is obvious when $f$ is an indicator function. Hence it is true for simple functions, hence for non-negative measurable functions and finally for all functions for which the integrals exist. For bounded continuous functions the integrals do exist.

expectation of random probability measure

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