This question was on math.stackexchange but got no answer. The link is <https://math.stackexchange.com/questions/598811/kolmogorov-backward-equation-for-ito-diffusion>


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?



Kolmogorov backward equation question
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.


