Suppose for i in [0,1], U(i) is a uniformly distributed independent random variable.Am I right in thinking that for any continuous function f:where E is the expectation operator? (That is if these integrals are actually defined?) Is there a reference for this?I imagine it's a straight forward application of the Monotone Convergence Theorem and the Law of Large Numbers, but I'm just a little bit worried by the fact that Brownian motion is often referred to as an integral of white noise, though perhaps more strictly it's an integral with respect to white noise, which is why that's different?Thanks in advance,Tom

I've just noticed there was a mistake in my description of the problem. f is also a function of i, so the actual question is whether:Thanks.

Hello Tom. I think that this equality can't hold in the general case.Notice that the integral in the lhs on your very last message it's a random variable and the integral on rhs is a number.Hope that help...

Well my intuition was that the integral on the left hand side was identical in all states of the world, by a law of large numbers style result and hence that it could be equal to the non-random right hand side.

Tom. What is the formal definition of the integral on the right hand side? Where did you saw that object?

On the left hand side I presume you mean? U(i) is the realisation of the (standard) uniform random variable, so it's just an integral over an everywhere discontinuous function. Treating it as a Lebesgue integral is fine. In the case when f is not a function of i, by the Strong Law of Large numbers, with probability 1 the integral just equals the expectation of f(U) where U is a standard uniform. (This is clearly true for the Riemann integral by considering Riemann sums and hence is also true for the Lebesgue.)Thinking about it, for it to generalise to the case in which f is a function of i, maybe I need to assume some kind of monotonicity relationship, e.g.:f(i,x) <= f(j,x) implies f(i,y) <= f(j,y) for all y in [0,1]Tom

yes. sorry. I meant the left hand side. I was thinking about the following counter-example:Take some omega such that U_i <= 0.5 if i \in V, and U_i > 0.5 if i \nin V, where V is the Vitali set. Also, fix f(x) = 0, if x <= 0.5 and f(x) = 1, if x > 0.5then, clearly f(U_i(w)) is not Lebesgue measurable.I guess (but I cant prove it) that the set of such omegas has probability > 0 (i think it could be probably 1)...that makes your integral some non well defined object since it is a random variable and as such, it has to be defined for every omega...When you talk about Riemann sums, keep in mind that f(U_i) is a random variable, and is a function on [0,1] for each fixed omega. but U_i is not continuous with probability 1, so you can't approximate its integral by Riemann sums, and therefore I don't think that your argument with the strong law holds...

On the first point we're just comparing guesses. My guess was that the probability of non-measurability was 0. Indeed it seems trivial to me that the integral of the function you describe is 0.5 with probability 1. But I may well be wrong.Your second point is worrying though. From the Wikipedia page the standard way of proving Riemann non-integrability seems to be to construct partition refinements going to different limits, and I guess for any omega you could certainly do this since in any open set there are going to be arbitrarily many arbitrarily high/low U(i)s. Damn it.At least with f not a function of i this really does get done all the time in Economics though. "Each firm in the continuum [0,1] has a 10% chance of updating its price in any period, therefore the aggregate price will be 0.1 * ... + 0.9 * ..." etc.It must be a different type of integral I guess...