Kolmogorov backward equation question

Question

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I want to know the answer too, so here it is

Let $(X_t)_{t\ge 0}$ be the solution of the SDE $$ X_t = X_0 + \int_0^t \mu(s,X_s) \,ds + \int_0^t \sigma(s,X_s) \,dB_s, \quad t\ge 0 $$ where $\mu(s,x)$ and $\sigma(s,x) $ are Lipschitz continuous in $x$ uniformly in $t$. My question is related to the last argument in a proof (in the book of Klebaner - Introduction to Stochastic Calculus in Section 6.2 on p.154) showing that $u(t,x):=\mathbb{E}(g(X_T)|X_t=x)$ solves $$ \frac{\partial}{\partial_t}u(t,x) + \mu(t,x)\frac{\partial}{\partial_x}u(t,x) + \frac{\sigma^2(t,x)}{2}\frac{\partial^2}{\partial x^2}u(t,x)=0.\qquad (*) $$ In the proof it is shown that $$ \int_0^t \frac{\partial}{\partial_t}u(s,X_s) + \mu(s,X_s)\frac{\partial}{\partial_x}u(s,X_s) + \frac{\sigma^2(s,X_s)}{2}\frac{\partial^2}{\partial x^2}u(s,X_s) \,ds = 0 \quad \mathbb{P}-a.s. $$ for every $t\ge 0$. Now it is concluded that the PDE $(*)$ above holds. Why is that?

Answer 1

First note the following fact: if $f : [0,T] \to \mathbb{R}$ is an integrable function and $\int_0^t f(s)\,ds = 0$ for all $0 \le t \le T$, then $f = 0$ almost everywhere. (It's immediate that $\int_a^b f(s)\,ds = 0$ for all $a,b$. Now use a monotone class argument to show that $\int_A f(s)\,ds = 0$ for every measurable $A$. Finally consider $A = \{f > 0\}$ and $A = \{f < 0\}$.)

So if we call the integrand $F(s,\omega)$, what we have shown is that for almost every $\omega$, $F(s,\omega) = 0$ for almost every $s$ (the null set of $s$ depending on $\omega$). But then Fubini's theorem implies that $F(s,\omega) = 0$ for almost every $(s,\omega)$ with respect to product measure.