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.