It is well known that if $\mu$ and $\nu$ are two measures on the space $C^0([0,1],\mathbb{R}^n)$ of continuous mappings from $[0,1]$ to $\mathbb{R^n}$ endowed with its Borel $\sigma$-algebra satisfy $$\int_{C^0([0,1],\mathbb{R}^n)} \left(\int_0^1 \phi(t,x(t))dt\right) d\mu(x(\cdot))=\int_{C^0([0,1],\mathbb{R}^n)} \left(\int_0^1 \phi(t,x(t))dt\right) d\nu(x(\cdot)),$$ for every measurable $\phi:[0,1]\times \mathbb{R}^n\to \mathbb{R}$, then we have $\mu=\nu$.
My question is:
- Do we still obtain $\mu=\nu$ if $\mu$ and $\nu$ are measures on the space $L^1([0,1],\mathbb{R}^n)$ (or by the set $M([0,1],X)$ of measurable functions having values in a compact set $X$ of $\mathbb{R}^n$) instead of $C^0([0,1],\mathbb{R}^n)$ ?
Any help is appreciated, thanks.
(For the continuous case, one can write $\phi(t,x)=\phi_1(t)\phi_2(x)$, then apply Fubini's Theorem and use the evaluation map and obtain the equality since it is easy to see that the two measures coincide on cylindrical sets.)
