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Let $\varphi :I\to J$ be a $C^1$ diffeomorphism between intervals, which may be unbounded. For a nonnegative Lebesgue measurable function $f$ on $J$, Lebesgue substitution on an interval givesallowing the value infinity. The same identity applied to $|f|$ shows that $f$ is Lebesgue integrable exactly when the transformed integrand is integrable; in that case the displayed formula holds for signed or complex $f$ as well. The absolute value on the derivative makes the formula valid for both increasing and decreasing substitutions. Finite endpoints can be omitted without changing a Lebesgue integral, because individual points are null sets.

Because $f$ is continuous, $g(x,y)=e^{-x^2}f(e^{-x^2/2}y)$ is continuous on ${\mathbb{R}}^2$, so $|g|$ is a nonnegative measurable function. Apply Tonelli's theorem before assuming integrability. For each fixed $x$, set $z=e^{-x^2/2}y$, so $dy=e^{x^2/2}dz$. Lebesgue substitution on an interval givesTherefore the Gaussian integral yieldsThis proves that $g$ is Lebesgue integrable on ${\mathbb{R}}^2$. Fubini's theorem now permits changing the order for $g$ itself. The same substitution without absolute values gives the stronger exact identityTaking absolute values and using the triangle inequality for an integral provesThe constant is sharp: any nonzero nonnegative continuous integrable $f$ gives equality. This is a weighted dilation kernel with $a(x)=e^{-x^2}$ and $b(x)=e^{-x^2/2}$. The inner $x$ integral in the requested order is guaranteed to be integrable for almost every $y$, which is all the iterated integral requires.

Let $f$ be a Borel measurable function in L1 space on $\mathbb{R}$, and let $a,b$ be measurable real functions of $x$ with $b(x)>0$ and $W={\int }_{\mathbb{R}}|a(x)|/b(x)dx<\infty$. The joint integrand $G(x,y)=a(x)f(b(x)y)$ is measurable, and Tonelli's theorem followed by Lebesgue substitution on an interval givesFor each fixed $x$, substitute $z=b(x)y$ in the inner $y$ integral. The absolute Jacobian is $1/b(x)$, leaving $|a(x)|\Vert f{\Vert }_1/b(x)$. Integrating in $x$ proves the formula and absolute integrability. Fubini's theorem then permits both orders of signed integration, givingThe one-dimensional sections are integrable for almost every value of the other variable; the theorem does not assert integrability of every section.