# Expectation of a generalized real-valued Brownian motion

Let $(B_t)_{t\ge 0}$ be a real-valued Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname P)$, $\lambda$ be the Lebesgue measure on $\mathbb R$ and $$\langle W,\phi\rangle:=\int_{[0,\infty)}\phi(t)B_t\;{\rm d}\lambda\;\;\;\text{for }\phi\in\mathcal D:=C_c^\infty(\mathbb R)\;.$$ Note that $\phi B(\omega,\;\cdot\;)$ is $\lambda$-integrable over $[0,\infty)$ for almost every $\omega\in\Omega$, since $B$ is almost surely continous and $\operatorname{supp}\phi$ is compact for all $\phi\in\mathcal D$.

I would like to change the order of integration in $$\operatorname E[W](\phi):=\operatorname E\left[\langle W,\phi\rangle\right]\;\;\;\text{for }\phi\in\mathcal D$$ in order to obtain (using $B_t\sim\mathcal N_{0,t}$) $$\operatorname E[W](\phi)=\int_{[0,\infty)}\phi(t)\underbrace{\operatorname E\left[B_t\right]}_{=0}\;{\rm d}\lambda=0\;\;\;\text{for all }\phi\in\mathcal D\;.$$

Let's assume that $B$ is (surely) continuous. Then (by continuity) $\phi B$ is $\operatorname P\otimes\lambda$-measurable and it remains to show that $\phi B\in\mathcal L^1(\operatorname P\otimes\lambda)$ for all $\phi\in\mathcal D$. After that we can change the order of integration by Fubini's theorem. So, how do we show that?

Can we obtain the result without the additional assumption of sure continuity, i.e. if $B$ is only almost surely continuous?

Since $\phi$ is continuous with compact support, it suffices to show that for each $R$, $$\mathbb E\left[\int_{[0,R]}|B_t |\mathrm d\lambda(t)\right]$$ is finite. By Schwarz inequality, it suffices to check finiteness of $$I:=\mathbb E\left[\int_{[0,R]}B_t^2\mathrm d\lambda(t)\right].$$ Using Fubini-Tonnelli theorem (that is, switching integral when the integrand is non-negative) and noticing that $\mathbb E\left[B_t^2\right]=t$, we get finiteness of $I$.
• I've edited question slightly cause I made a mistake. I've said that $\phi B$ would be product measurable by continuity, but that's not true cause $B$ is only almost surely continuous. We need the product measurability in order to apply Fubini's theorem. Can we prove the result without the additional assumption (made in the question)? I don't think so. – 0xbadf00d Jan 30 '16 at 18:43