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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?