Measure Theory/Outer Measuring Intervals

Outer Measuring Intervals

Exercise 1. Singleton Sets Are Null 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 $\lambda ^{*}([a,b])\leq b-a$.

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

The corresponding over-estimate is $b-a+2\varepsilon$.

Exercise 2. Carry the $\lambda ^{*}([a,b])\leq b-a$ Complete the argument that, due to what we have seen above, $\lambda ^{*}([a,b])\leq b-a$.

In order to show $b-a\leq \lambda ^{*}([a,b])$, 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 ${\mathfrak {O}}$ be any open interval over-approximation of $[a,b]$. Let $e=\sum _{I\in {\mathfrak {O}}}\ell (I)$ be the corresponding over-estimate.

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

Notice that every term which makes up e' also occurs in e. Therefore $e'\leq e$.

Further, it will be useful if no interval in ${\mathfrak {O}}'$ is a subset of any other interval in it. Therefore we may construct a still smaller approximation, ${\mathfrak {O}}''$ which results from removing intervals from ${\mathfrak {O}}'$ until no interval that remains is a subset of any other interval in ${\mathfrak {O}}''$.

Exercise 3. Removing Subset Intervals Is Fine Prove that ${\mathfrak {O}}''$ constructed above is still an open interval over-approximation of $[a,b]$.

Let $e''$ be the over-estimate corresponding to ${\mathfrak {O}}''$. As before, we have $e''\leq e'$.

Now assume that ${\mathfrak {O}}''=\{(a_{1},b_{1}),(a_{2},b_{2}),\dots ,(a_{n},b_{n})\}$ and assume that these intervals are listed "in order". Here "in order" means that $a_{1}<a_{2}<\cdots <a_{n}$.

Exercise 4. Check the Interval Order 1. Prove that, with the ordering given above for the intervals, $a_{i}<b_{i-1}$ for each $2\leq i\leq n$. Hint: If this were not true, you would get an interval inside another interval in ${\mathfrak {O}}''$. 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 $e''\geq b-a$. Reason through the remainder of the proof.

Exercise 5. Outer Measure of Other Intervals 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 $[a+\varepsilon ,b-\varepsilon ]$ and then use monotonicity to prove $b-a-2\varepsilon \leq \lambda ^{*}((a,b))$. Then let $\varepsilon \to 0$. The reverse inequality should be even more direct. The use of monotonicity and the outer measure of closed intervals, gives an easy proof that $\lambda ^{*}(I)=\infty$ if I is an unbounded interval.