Lagmental Vicfred

Fubini Requires Absolute Integrability by Vicfred

An integrable signed or complex function has almost-everywhere integrable slices and equal iterated integrals. I want the notation, the mechanism, and the failure mode visible at the same time.

Notation

Lebesgue integration treats a measurable function \(f:X\to[-\infty,\infty]\) through level sets and simple approximations. The space \(L^1(\mu)\) consists of integrable functions modulo equality almost everywhere.

$$ \int_{X\times Y}|f|\,d(\mu\times\nu)<\infty $$

I keep the defining relation \(\mathsf D\) above the derived relation \(\mathsf C\). This exposes whether cancellation used \(x\ne0\) and whether the conclusion is canonical.

$$ \int_{X\times Y}f=\int_X\int_Yf=\int_Y\int_Xf $$

Stress the formula

A worked instance is useful here because it exposes every index that the compressed statement hides.

$$ \left|\int_X\!\int_Yf(x,y)\,d\nu\,d\mu\right|\le\int_X\!\int_Y|f(x,y)|\,d\nu\,d\mu<\infty $$

Interpretation

What survives the example is not its particular numbers but the relation encoded by the two rows below. That relation is the part worth transporting to a new setting.

$$ \begin{aligned} \mathsf{D}\;&:\quad \int_{X\times Y}|f|\,d(\mu\times\nu)<\infty,\\[5pt] \mathsf{C}\;&:\quad \int_{X\times Y}f=\int_X\int_Yf=\int_Y\int_Xf. \end{aligned} $$

Limit of the argument

Every convergence theorem has a different hypothesis. Pointwise convergence alone does not permit an integral and a limit to change places.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \int_{X\times Y}f=\int_X\int_Yf=\int_Y\int_Xf \end{gathered}} $$

The important habit is to remember what was fixed before the calculation began and what was proved only afterward. The final display preserves that order.

This article was posted on Sat 15 December 2012. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.