The Fundamental Theorem of Calculus states that if $f$ is continuous on $[a,b]$, then if we take $F(x)=\int_a^x f(t)dt$ for $x\in [a,b]$, we have $F'(x)=f(x)$ for all $x\in (a,b)$.
My question is how does this also apply in the case where we have $a=-\infty$ or $b=\infty$.
I am asking this question because, for example, when we compute the density of the pair $(M_t,B_t)$ where $M_t$ is $\sup_{s\le t} B_t$ and $B_t$ is a Brownian motion, we start with the fact that $$P[M_t\ge y, B_t \le x] = P[B_t \ge 2y-x]=\int_{2y-x}^\infty (2\pi t)^{-1/2} e^{-z^2/(2t)}dz$$ and then set $F(y,x)= P[M_t\le y, B_t \le x] = P[B_t \le x] - P[M_t \ge y, B_t \le x]$. So we have $$f(x,y)=-\frac{\partial^2}{\partial x \partial y} P[M_t \ge y, B_t \le x] = \frac{2(2y-x)}{\sqrt{2\pi t^3}} e^{-(2y-x)^2/(2t)}dxdy. $$
But this final equality is achieved by first differentiating $u\mapsto \int_u^\infty (2\pi t){-1/2} e^{-z^2/(2t)}dz$ and applying chain rule to the composition $(y,x)\mapsto u$. So the first differentiation is of the form given in the statement of the FTC, but the upper limit is $\infty$. I have not seen a mathematical justification of this before, and I would appreciate if anyone could provide an explanation to this.