Measure Theory, Integrable Function

Integrable Function

In simple terms, the integral of a function f is the integral of the positive part, minus the integral of - the negative part. This builds on what we already know. We defined the integral for a simple function, then a nonnegative function, and now a wide class of functions that don't stray off into infinity.

This definition stands on its own, but most textbooks require f to be measurable to be "integrable". If f isn't measurable, then we have to take extra steps to make sure the regions of X, where f is positive and negative, are measurable sets in X. Otherwise we can't integrate each piece separately, and the definition falls apart. So let's follow the crowd, and assume f is measurable, which is a prerequisite for integrable.

Let f⁺ equal f whenever f is positive, and 0 elsewhere. To prove this is measurable, we need our usual assumption about the topology of R. The open ray, starting at 0 and going up, is measurable, and so is its preimage under f. This is the part of X that maps into positive reals or +∞. Combine this with 0 on the rest of X and find a measurable function by concatenation. Similarly, f^- = -f where f was negative, and 0 otherwise. This too is measurable by concatenation. Finally, f⁰ = f where f is 0, and 0 otherwise; but this is the same as 0.

Note that f = f⁺-f^-. Also, |f| = f⁺+f^-.

Let L(X,σ,μ) denote the measurable functions from X into the extended reals, such that ∫ f⁺ and ∫ f^- are finite. These functions are all integrable by definition, and the integral is as follows.

∫ f = ∫ f⁺ - ∫ f^-

Note that the integral of an integrable function is finite.

If f meets both criteria simultaneously, i.e. nonnegative, and measurable, and its positive and negative portions have finite integrals, then the integral of f is, well, the integral of f. Where there is overlap, this definition is consistent with the one we established earlier, and there is no ambiguity.

Concatenation

Partition X into a countable collection of disjoint measurable regions, and the integral of f over x is the sum of the integral of f over the individual regions. Concatenation holds for f⁺, and for f^-, so put these together and regroup terms to get the desired result. If the number of regions is infinite, we're still all right, because the series are absolutely convergent.

Linear Combinations of Functions

Show that the integral of k×f is k times the integral of f. It's true for f⁺, and f^-, so put these pieces together and it's true for f. Thus scaling and integration commute.

A finite set of functions partitions X into a finite number of regions, such that each function, on a given region, is positive, negative, or 0. Within each region, the integral of the linear combination of the sign-adjusted functions is the linear combination of the integrals. Put this all together, managing the signs properly, and integration is a linear operator, as we expect.