Let $p(u,x):=(4 \pi u)^{-1/2}e^{-\frac{x^2}{4u}},u>0,x \in \mathbb{R}.$ Let $\mathcal{P} \subset \mathcal{B}(\mathbb{R}_+) \otimes \mathcal{F}$ the predictable $\sigma$-algebra. Let $u_0:\mathbb{R} \times \Omega \to \mathbb{R}$ be $\mathcal{B}(\mathbb{R}) \otimes \mathcal{F}_0$-measurable and $\sup_{x \in \mathbb{R}}E[(u_0(x))^2]<+\infty.$
We want to prove that $\int_{\mathbb{R}}p(u,x-y)u_0(y,\omega)dy,u>0,x \in \mathbb{R},\omega \in \Omega$ is well-defined and $\mathcal{P} \otimes \mathcal{B}(\mathbb{R})$-measurable.
While I was reading in a document, they proved for, fixed $u>0,x \in \mathbb{R},$ that $E\left[\left|\int_{\mathbb{R}}p(u,x-y)u_0(y)dy\right|^2\right] \leq \sup_{x \in \mathbb{R}}E[(u_0(x))^2],$ can you make sense of $\int_{\mathbb{R}}p(u,x-y)u_0(y)dy$ using this inequality, in other words it's defined on an event $F_{u,x}$ such that $P(F_{u,x})=1$? Can you obtain an event $F$ independent of $u,x$? What about the $\mathcal{P} \otimes \mathcal{B}(\mathbb{R})$-measurability?
For a bounded measurable $I \subset \mathbb R^+ \times \mathbb R$, we have \begin{align*} &\mathbb \int_I \int_\mathbb R \mathbb E[p(u, x-y) \lvert u_0(y)\rvert]\ dy\ dx\ du\\ &\qquad \le \sup_{y\in\mathbb R} \mathbb E[u_0(y)^2]^{1/2} \int_I \int_\mathbb R p(u, x-y) \ dy\ dx\ du\\ &\qquad < \infty. \end{align*} From Fubini's theorem, we deduce that the integrals $$ v_0(u, x) = \int_\mathbb R p(u, x-y) u_0(y) \ dy $$ are well-defined almost-everywhere on $I \times \Omega$, and $v_0\mid_I$ is $(\mathcal B(I) \otimes \mathcal F_0)$-measurable. By the continuity of $p$, we can further obtain that $v_0$ is well-defined everywhere on $I$ almost-surely.
As we can cover $\mathbb R^+ \times \mathbb R$ with countably many such $I$, we thus have that $v_0$ is well-defined everywhere almost-surely, and is $(\mathcal B(\mathbb R^+ \times \mathbb R) \otimes \mathcal F_0)$-measurable, which implies the result.