# Integral of a Continuum of i.i.d. Random Variables

Fix some measurable space, $$(\Omega,\mathcal{F})$$ Suppose that we have a continuum of i.i.d. random variables $$\{X_i\}_{i\in[0,1]}$$ distributed according to cdf, $$G(\cdot)$$ (denote the corresponding pdf by $$g(\cdot)$$). Define $$X:=\int_0^1X_i(\cdot)d\lambda(i)$$, where $$\lambda(\cdot)$$ is the Lebesgue measure on $$\big([0,1],\mathcal{B}([0,1])\big)$$.

Question: What is the distribution of X?

Notes:

• For simplicity, I assume that the $$G$$ (and $$g$$) are either only continuous or only discrete probability distributions (eg, all of the $$X_i$$'s are iid $$\mathcal{U}([0,1])$$ rv's). If there is a particular $$G$$ that is easy for you to explain, please feel free to suggest one.

• I'm not sure such an integral is well defined. If $X_i$ are Bernoulli 0/1 then why should we expect the set of all $i \in [0,1]$ for which $X_i=1$ to be measurable? The only case I can see for which the integral makes sense is if all random variables are (surely) the same constant $c$. Jun 11 '19 at 3:48
If $$(\omega,i)\mapsto X_i(\omega)$$ were jointly measurable (and bounded, say), then you could (i) do the integration, and (ii) $$X$$ so defined would be a random variable. But this can't be unless the $$X_i$$ are degenerate. To see this, assume the joint measurability (on top of the iid) and compute the variance of $$X$$. You'll get $$0$$, which will force you to conclude that the $$X_i$$ were degenerate all along.