The plane Gaussian integral equals $\pi$ by polar coordinates

On each compact rectangle $[\epsilon ,R]\times [0,\pi ]$ and $[\epsilon ,R]\times [-\pi ,0]$, the polar map has Jacobian determinant $r({\cos }^2\theta +{\sin }^2\theta )=r>0$ by [L12], [L13], and [L17]. Enlarge the radial interval inside $(0,\mathrm{\infty })$ and each angular interval by less than $\pi \mathrm{/}2$. If two points in one enlarged rectangle have the same polar image, [L13] gives equal positive radii; reducing both angles modulo $2\pi$ by [L23] and using its injective half-open parametrization shows that their difference is an integer multiple of $2\pi$. The enlarged angular interval has length below $2\pi$, so the angles are equal. Thus the map is injective on an open neighbourhood of each compact rectangle, and their images are the closed upper and lower half-annuli. Each half-annulus is closed and bounded, hence compact by [L6]; its boundary is contained in two continuous semicircle graphs and two radial segments, so [L4] and [L5] make it Jordan measurable.

The Gaussian is continuous on each compact half-annulus and the pulled-back function $e^{-r^2}r$ is continuous on each parameter rectangle, so [L22] supplies both integrability conditions in [L1]. Apply [L1] to each half-annulus. By [L18], the parameter-rectangle integral is $\pi {\int }_{\epsilon }^R{e}^{-r^2}rdr$. Facts [L10], [L19], and [L20] give $(-\frac{1}{2}{e}^{-r^2}{)}^{\mathrm{\prime }}=r{e}^{-r^2}$, so [L14] evaluates each half as $\frac{\pi }{2}(e^{-{\epsilon }^2}-e^{-R^2})$; the lower half has the same angular length.

The half-annuli overlap only in the two radial boundary segments, which have content zero by [L4]. On a common bounding rectangle, $1_{A_{+}}+1_{A_{-}}$ differs from $1_{A_{+}\cup A_{-}}$ only on that overlap, so [L3], [L7], and [L8] combine the two values from step 2.1 into the full-annulus integral $\pi (e^{-{\epsilon }^2}-e^{-R^2})$.

Every circle is the union of its upper and lower continuous semicircle graphs, hence has content zero by [L4]; [L5] makes every closed disc Jordan measurable, and [L6] makes it compact.

The omitted inner disc lies in $[-\epsilon ,\epsilon {]}^2$, whose content is $4{\epsilon }^2$ by [L7] and [L9]. Since $-(x^2+y^2)\le 0$, strict increase in [L16] and the normalization/positivity in [L21] give $0<e^{-(x^2+y^2)}\le 1$. Thus [L8] bounds its Gaussian integral by $4{\epsilon }^2$. The annulus and inner disc overlap only on a content-zero circle, so [L3] recombines them. Letting $\epsilon \downarrow 0$ in step 3.1 gives the radius-$R$ disc integral $\pi (1-e^{-R^2})$.

The closed discs of radii $j+1$ form a compact Jordan exhaustion of ${\mathbb{R}}^2$. The Gaussian is nonnegative by [L21], and it is locally Riemann integrable because [L10] and elementary algebra make it continuous and [L22] makes its restriction to every compact Jordan set integrable. Thus [L11], step 5.1, and [L15] show that the disc integrals tend to $\pi$, which is the improper plane integral.