# 16.4 Lebesgue Integration

The Riemann integral with integrand $$f(x)$$ over a given interval is defined as the limit as $$d$$ approaches $$0$$ of the sum over subintervals of size $$d$$, of $$d$$ multiplied by the value of $$f$$ at an arbitrary point in that subinterval, when that limit exists and is the same for every choice of points in all the intervals. Otherwise the function is said to be not integrable over that interval.

A function that is $$1$$ on all rational numbers and $$0$$ on all other numbers is obviously not integrable in this sense, since every interval of any non-zero size contains both rational and irrational numbers and hence values for this function of both $$0$$ and $$1$$. There are many more irrational numbers than rational ones, which suggests the possibility that we could just forget about the rational ones and say that the integral is $$0$$. On the other hand, if we perform a numerical computation on a computer, since the computer rounds off every point to a rational one, we would find the value $$1$$ for every interval.

There is another way to define the integral of a function for which the function just described is integrable.

Instead of splitting the area computed by the integral by slices parallel to the $$y$$ axis, we can split into pieces by slices parallel to the $$x$$ axis.

Suppose, for convenience that the integrand function is non-negative and we integrate over a finite interval. Then for each slice we will find that for some points within the domain the slice is below the integrand, for some it is above it, and for some the integrand lies within the slice. As the size of the slices decreases, the contributions from the last of these becomes negligible and the integral will be the sum of the contributions for which the slices are below the integrand.

For a continuous integrand, for each slice the points below it will form some set of intervals on the real line. There are many kinds of integrals as we shall soon notice. In each we define a "measure" to each slice of the point set for which the slice is below the integrand, and the sum of these measures over all slices must converge to the integral.

What constitutes a measure? The main necessary condition is the sum of the measures of disjoint sets (these are sets having no common point) is the measure of their union (the union being the set of points that are in either one). This must apply to the union of any finite number of mutually disjoint sets: their union must have measure equal to the sum of their measures. Since any point in a countable list of points is in a finite initial segment of the countable list, it makes sense that the measure of a union of any countable number of sets must be the sum of their measures.

In the case of the usual integral $$\int f(x)dx$$, the measure of an interval is its length. A single point is an interval of length $$0$$, and has $$0$$ measure.

This tells us that the measure of any countable set of points must be a countable sum of $$0$$'s and hence must be $$0$$ as well with this measure (and measures like them.)

We can conclude that for the usual integral, that weird function that is $$0$$ on rational numbers and $$1$$ on the rest of the real numbers, is integrable. The rational numbers, being countable, have measure that is a countable sum of $$0$$'s and hence is $$0$$. Thus the measure of the rest is the length of the interval integrated over, and this weird function is indeed integrable.

What other measures are there?

We have encountered some other measures already; if we deal with $$\int f(x)dy(x)$$, which is $$\int f(x)y'(x)dx$$, we are integrating $$f(x)$$ using a measure defined by $$y'(x)$$. For example, if $$y(x)$$ is $$x^2$$, the measure of an interval is not its length, but the difference of $$x^2$$ between its values at the endpoints of the interval.

Also, ordinary sums, for example $$f(x_1) + f(x_2) + f(x_3)$$, can also be written as Lebesgue integrals, in this case using the measure that is $$1$$ on the points $$x_1, x_2$$, and $$x_3$$, and $$0$$ elsewhere.

That sort of integral has been used since the $$19^{th}$$ Century by physicists, and for some time was frowned upon by mathematicians. Physicists introduced the "delta function" $$\delta(x-x_0)$$ which is $$0$$ unless $$x$$ is $$x_0$$, but its integral is $$1$$. The integral giving the sum of the previous paragraph can then be written as

$\int dx f(x)(\delta(x-x_1) + \delta(x-x_2) + \delta(x-x_3))$

The obvious problem with the delta function $$\delta(x-x_0)$$ is that it would have to be infinite when $$x$$ is $$x_0$$. Fortunately, the consequences of allowing it to have an unmeasurably small width around $$x_0$$ as a function of $$x$$, in which case it can remain finite, are undetectable for all uses in which, when it is ultimately applied, it is integrated over. And from the point of view of Lebesgue integration, it is just another measure.