Measure Theory/Length Measure

Length Measure
We now have a good introductory comprehension of the outer measure, $$\lambda^*$$. What we wanted this whole time, though, was the measure function, which we'll call $$\lambda$$. Finally now we are in a position to state its definition.

Recall that in previous lessons we stated that $$\lambda$$ will simply be the same thing as $$\lambda^*$$, except with its domain restricted to "nice" or "not weird" sets. We can nearly say what the nice sets are, and therefore specify the domain restriction.

But first we must introduce the inner measure.

Inner Measure
You have probably noticed a somewhat curious asymmetry in the definition of the outer measure. Why take over-approximations? Could we have taken under-approximations instead?

Indeed we could have. For reasons that are a bit too much of a distraction, if we were to approximate sets "from below" we would need to use closed sets instead of open sets.

This is exactly what the inner measure does.

We could, but will not, go through all the same sorts of theorems for inner measure which we just finished for outer measure. Rather, we will simply take some results about inner measure for granted. The reader is free to research the proofs of these facts independently.

Measurable Sets
The strategy for defining the measurable sets is extremely similar to the way in which we define integrable functions in elementary analysis.

Now since we are not going to seriously study the inner measure, I need to give you another way to tell whether a set is measurable or not.

Indeed, there is a much better way than checking whether the definition holds.

We will now state and use, without a proof, the following theorem:

Let $$E\subseteq\Bbb R$$ be a set of real numbers. Then E is measurable if and only if E splits every set cleanly.

The condition that a set splits every set cleanly, is called the "Caratheodory condition" in honor of its discoverer.

Prove that the empty set is measurable, $$\emptyset\in\mathcal M$$.

Prove that every null set is measurable. That is to say, if $$N\subseteq\Bbb R$$ is a null set, then $$N\in\mathcal M$$.

Let $$E\in\mathcal M$$ be any measurable set. Prove that the complement is also measurable, $$E^c \in\mathcal M$$.

Open Rays Are Measurable
Since the whole point of the project is to measure intervals and interval-like-things, we would like to prove that intervals are measurable.

The proof is not very easy though. One way to simplify the proof is to consider only intervals of the form $$(a,\infty)$$, which we call a "ray" of real numbers. The fact that it only has one end-point makes it a little bit easier to use than a bounded interval.

We now set for ourselves the proof of the following theorem.

Let $$a\in\Bbb R$$. Then the open right ray $$(a,\infty)\subseteq\Bbb R$$ is measurable, $$(a,\infty)\in\mathcal M$$.

Let $$I = (a,\infty)$$, and let $$A\subseteq\Bbb R$$ be arbitrary. We need to show that I splits A cleanly.

As is typical for proofs in analysis, we will approach the problem by proving two inequalities.


 * $$\lambda^*(A)\le \lambda^*(A\cap I) + \lambda^*(A\smallsetminus I)$$

and


 * $$\lambda^*(A)\ge \lambda^*(A\cap I)+\lambda^*(A\smallsetminus I)$$

But notice that in fact the first inequality follows immediately from subadditivity. In fact, whenever proving that a set splits another cleanly, we can always use subadditivity for the $$\le$$ direction.

Therefore we only need to prove the second inequality.

Consider the case that $$\lambda^*(A)=\infty$$ and state why, in this case, there is nothing to prove.

Due to the result above, we will assume throughout the rest of the proof that $$\lambda^*(A)$$ is finite.

We will initially assume $$a\not\in A$$. This is simplifying, and once we have proved our theorem in this case, we can use the narrow result to prove the theorem in the case where $$a\in A$$.

The high-level strategy of our proof will be to show that $$\lambda^*(A\cap I)+\lambda^*(A\smallsetminus I)$$ is a lower bound on the set of over-estimates of A. From this, the needed inequality follows immediately.

To begin, let $$\varepsilon\in\Bbb R^+$$ and let $$\mathfrak O$$ be open interval over-approximations of A such that


 * $$ \sum_{J\in\mathfrak O}\ell(J)<\lambda^*(A)+\varepsilon $$

Show that $$\mathfrak P = \{U\cap I|U\in\mathfrak O\}$$ is an open interval over-approximation of $$A\cap I$$.

Also do similarly for $$A\smallsetminus I$$. When you decide what the open interval over-approximation of $$A\smallsetminus I$$ should be, call it $$\mathfrak Q$$.

Now show that, for any $$U\in\mathfrak O$$,


 * $$\ell(U) = \ell(U\cap I)+\ell(U\smallsetminus I)$$

Hint: U cannot be infinite in length, otherwise the over-estimate above is infinite, in which case it cannot be strictly bounded by any number.

Therefore $$U = (b,c)$$ for finite real numbers $$b,c\in\Bbb R$$.

Justify why


 * $$\lambda^*(A\cap I)+\lambda^*(A\smallsetminus I) \le \sum_{J\in\mathfrak P}\ell(J)+\sum_{K\in\mathfrak Q}\ell(K) $$

and why the right-hand side equals


 * $$\sum_{L\in\mathfrak O}\ell(L)$$

which is less than


 * $$\lambda^*(A)+\varepsilon$$

Then conclude the proof that $$\lambda^*(A)\ge \lambda^*(A\cap I)+\lambda^*(A\smallsetminus I) $$.

Finally, prove $$\lambda^*(A)\ge \lambda^*(A\cap I)+\lambda(A\smallsetminus I)$$ in the case that $$a\in A$$.

Hint: By an earlier exercise, you can show all three of


 * $$\lambda^*(A) = \lambda^*(A\smallsetminus \{a\})$$
 * $$ \lambda^*(A\cap I) = \lambda^*([A\smallsetminus \{a\}]\cap I]) $$
 * $$ \lambda^*(A\smallsetminus I) = \lambda^*([A\smallsetminus \{a\}]\smallsetminus I]) $$

Now apply the result of ''Exercise 5. Using the Split Covers'' to the set $$A\smallsetminus \{a\}$$.

Length Measure
Note that the only way in which $$\lambda$$ and $$\lambda^*$$ differ is by their domain. The former has domain $$\mathcal M$$ while the latter has $$\mathcal P(\Bbb R)$$.

Whenever the set E is measurable we will prefer to write $$\lambda(E)$$ instead of $$\lambda^*(E)$$ even though they technically mean the same thing. When a set is not necessarily measurable we will be required to write $$\lambda^*(A)$$.

Therefore we are now justified in writing, for example,


 * $$\lambda(\emptyset) = 0$$
 * $$\lambda(\Bbb Q) = 0$$
 * $$\lambda((0,\infty)) = \infty$$

On the other hand, if F is the feather set from the Vitali proof that there is no total measure function, then one would have to write $$\lambda^*(F)$$ since we have no current reason to think that it is measurable.