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 $C^0([0,1],\mathbb{R}^n)$ ?

(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.)

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.)