Measure Theory/Outer Measuring Intervals

Outer Measuring Intervals

Template:Robelbox Prove that any singleton set is a null set.

Hint: You may prove this directly as an exercise, but it also follows very trivially from a previous exercise.

[a,b]

Now we show that for any closed, bounded interval, ${\lambda }^{*}([a,b])=b-a$. As is typical with proofs in analysis, we do this by showing two inequalities.

The first inequality that we will show is ${\lambda }^{*}([a,b])\le b-a$.

We can accomplish this by observing that for any $\epsilon \in {ℝ}^{+}$ the set $\{(a-\epsilon ,b+\epsilon )\}$ is always an open interval over-approximation.

The corresponding over-estimate is $b-a+2\epsilon$. Template:Robelbox Complete the argument that, due to what we have seen above, ${\lambda }^{*}([a,b])\le b-a$.

In order to show $b-a\le {\lambda }^{*}([a,b])$, we will show that $b-a$ is a lower bound on the set of over-estimates. Recall that, by definition, ${\lambda }^{*}([a,b])$ is the greatest of all such lower bounds.

To this end we let $𝔒$ be any open interval over-approximation of $[a,b]$. Let $e=\sum _{I\in 𝔒}\ell (I)$ be the corresponding over-estimate.

Because $[a,b]$ is compact, and because $𝔒$ is an open cover, then there must exist a finite subcover, ${𝔒}^{\prime }\subseteq 𝔒$. Let $e^{\prime }$ be its corresponding over-estimate.

Notice that every term which makes up e' also occurs in e. Therefore $e^{\prime }\le e$.

Further, it will be useful if no interval in ${𝔒}^{\prime }$ is a subset of any other interval in it. Therefore we may construct a still smaller approximation, ${𝔒}^{″}$ which results from removing intervals from ${𝔒}^{\prime }$ until no interval that remains is a subset of any other interval in ${𝔒}^{″}$. Template:Robelbox Prove that ${𝔒}^{″}$ constructed above is still an open interval over-approximation of $[a,b]$.

Template:Robelbox/close Let $e^{″}$ be the over-estimate corresponding to ${𝔒}^{″}$. As before, we have $e^{″}\le e^{\prime }$.

Now assume that ${𝔒}^{″}=\{(a_1,b_1),(a_2,b_2),\dots ,(a_n,b_n)\}$ and assume that these intervals are listed "in order". Here "in order" means that $a_1<a_2<\cdots <a_n$. Template:Robelbox

1. Prove that, with the ordering given above for the intervals, $a_i<b_{i-1}$ for each $2\le i\le n$.  Hint: If this were not true, you would get an interval inside another interval in ${𝔒}^{″}$.

Further show that $a_1<a$ and $b<b_n$.  

2. Show that $e^{″}=b_n-a_1+\sum _{i=2}^n(b_{i-1}-a_i)$.  Hint: Write out the definition of $e^{″}$ as a sum and rearrange the terms.  It may help to not use sigma-notation for the summation, in order to see how to do the rearrangement.  

Infer, with the help of (1.), that $e^{″}\ge b-a$.

Reason through the remainder of the proof.

Template:Robelbox/close Template:Robelbox Use the result above to prove that the outer measure of any open interval also equals its length.

Hint: Start with a bounded open interval, (a,b). Approximate this "from below" with intervals of the form $[a+\epsilon ,b-\epsilon ]$ and then use monotonicity to prove $b-a-2\epsilon \le {\lambda }^{*}((a,b))$. Then let $\epsilon \to 0$.

The reverse inequality should be even more direct.

The use of monotonicity and the outer measure of closed intervals, gives an easy proof that ${\lambda }^{*}(I)=\mathrm{\infty }$ if I is an unbounded interval. Template:Robelbox/close