# Lebesgue Theory Part 2

Part 1 of this post focused on an introduction into sets, what a ring is, what a sigma-ring is, what does it mean to be additive, independent, and countably additive. We finished with a theorem that addressed the essence of countable additivity; this theorem set up the needed tools to evaluate some important aspect of a measure.

### MEASURE THEORY

#### DEFINITIONS AND PROPERTIES

Let us consider a set of points of (a, b); one can enclose in an infinite number of ways these points in an infinite number of intervals; the infimum of the sum of the lengths of the intervals is the measure of the set. A set E is said to be measurable if its measure together with that of the set of points not forming E gives the measure of (a, b).

He continues explaining,

Enclose A in a finite or denumerably infinite number of intervals, and let l1, l2, . . . be the length of these intervals. We obviously wish to have

m(A) ≤ l1 + l2 + · · ·

If we look for the greatest lower bound of the second member for all possible systems of intervals that cover A, this bound will be an upper bound of m(A). For this reason we represent it by m∗(A), and we have

m(A) ≤ m∗(A) (3)

If (a, b) \ A is the set of points of the interval (a, b) that do not belong to A, we have similarly

m((a,b) \ A) ≤ m∗((a,b) \ A).

Now we certainly wish to have

m(A) + m((a, b) \ A) = m(a, b) = b − a (4)

and hence we must have

m(A) ≥ b − a − m∗ ((a, b) \ A)

The inequalities (3) and (4) give us upper and lower bounds for m(A). One can easily see that these two inequalities are never contradictory. When the lower and upper bounds for A are equal, m(A) is defined, and we say that A is measurable.

What does Lebesgue mean by this? What is he getting at? Well, let us consider some set $A\subseteq\mathbb{R}$ And, for now, let’s not consider if $A\subseteq (a-b)$. Think about what his definition is for m*(A). Basically, let us cover the set A by countably many intervals {I},

the moment on whether A ⊆ (a, b); for example, A could be unbounded. Consider

his definition of m∗(A). Let’s cover A by countably many intervals ${I_n }$, so that $A\subseteq\bigcup\limits_{n=1}^{\infty} I_n$

For our purposes, we are going to (in the name of simplicity) assume that all our intervals $I_n\in\mathbb{R^1}$ , and note that the term $m(I)$ denotes the usual length of ${I}$. Now since $A\subseteq\bigcup\limits_{n=1}^{\infty} I_n$, the sum $\sum\limits_{n=1}^{\infty}m(I_n)$ should be larger than the “true measure” of our set A.  The more accurately the union of the sequences of I approximates A, the smaller our aforementioned sum is. The opposite is also true (i.e. if our countable union does not approximate the set well, our sum is going to be larger). The leftmost approximates poorly, and the furthest to there right approximates the set A closer. So, if we summed the “area” of the squares, we see, clearly, that the poor approximation is a larger value.

As Lebesgue did, we call m*(I) to be an upper bound of m(I), but not just any bound -it is the least upper bound. We define,

m*(A)= $inf\{\sum\limits_{n=1}^{\infty}m(I_n)\}$

We define this as the outer measure of A. The outer measure basically assigns a “length” to every subset of the reals -that is all!

In the next post, we will provide some information on elementary sets, what they do, and why we should care about them. Finally, we will relate this to our m(I), measure, and see how it pushes our definition forward.