# 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.