 Solution

The annulus is open and hence Lebesgue measurable; multiplying its indicator by the continuous Gaussian gives a measurable function. It is bounded by the indicator of $[-b,b{]}^2$, whose Lebesgue integral is $4b^2$. Thus it is Lebesgue integrable by comparison.

Use the change of variables formula with polar coordinates $x=r\cos \theta$, $y=r\sin \theta$. Removing one radial cut makes this a smooth one-to-one coordinate map; the cut is a null set and does not affect the integral. The absolute Jacobian determinant is $r$. By Fubini theorem,The antiderivative is $-e^{-r^2/2}$, which also makes the sign of the answer transparent.