Defining the integral of a function using the product measure

Question

Imagine that we're trying to define the expression $$\int_U f(x)dx$$ in a rigorous way. Assume that $f:X \rightarrow \mathbb{R}^{\geq 0}$ where $(X,\mu)$ is a measure space, and suppose that $U$ is a measurable subset of $X$. That most typical approach to making this integral rigorous is the method of Lebesgue, whereby we partition the range of $f$ into increasingly small horizontal strips. This seems very elaborate to me - why not just define the integral in the obvious way as the "(product) measure of the set of all points under the curve"? (if its defined; our integrable functions would then be precisely those for which the product measure is indeed defined). We can make this idea precise by writing

$$\int_U f(x)dx := (\mu \times \lambda)(\lbrace (x,y) : x \in U \wedge 0 \leq y \leq f(x)\rbrace)$$

where $\mu$ is the measure on $X$ and $\lambda$ is the standard measure on $\mathbb{R}$.

My question is, why isn't this the "standard" definition of the integral?

Answer 1

Linearity of this integral is very mysterious. Moreover, the definition of the product measure using integration, i.e. $\mu \otimes \lambda (M) =\int \int I_M(x,y) d\mu(x) d\lambda(y)$, is very easy (up to a technical problem concerning measurability) and can be understood without knowing Caratheodory's construction of measures.

Answer 2

I'm not sure I buy Jochen's comment that product measure can be so easily defined using integration --- it seems like you're going to have to do some work to show that his double integral is well-defined for every set $M$ in the $\sigma$-algebra generated by the measurable rectangles.

The real problem may be that you actually "need" integration theory to define product measures via the standard Caratheodory construction, when you show that $(\mu\times\nu)(A \times B) = \mu(A)\nu(B)$ defines a premeasure on the algebra generated by the measurable rectangles. That is, if $A \times B$ can be expressed as a disjoint union $\bigcup A_i \times B_i$, we need $\mu(A)\nu(B) = \sum \mu(A_i)\nu(B_i)$. And as far as I can see you pretty much have to use the monotone convergence theorem to prove that.

Answer 3

It is true that there is usually some redundancy in the treatment of measure and integration. One goes through two similar extension procedures, one for integration---from a simple family of functions to a more complicated one, the other for the measure from a simple family of sets to a more complicated one. Famously, in for Lebesgue theory say on euclidean space, from step functions to positive measurable ones and then to integrable ones in the first case and from intervals to measurable ones in the second. The two approaches are in a certain sense equivalent. If you can integrate functions you can define measurable sets and their measures via the indicator function and, as you point out, one can detect integrable functions (say positive) via the epigraph. Nevertheless, I think that there are very good pedagogical reasons for carrying out the two approaches independently. Another justificaiom lies in the fact that, as indicated in the answers and comments above, the transition between the two approaches is not quite as straightforward as it might appear at a first glance. Might I suggest as a useful compromise to bring both methods in class and use the above equivalence as a source of useful and illuminating exercises for students.