Measure Theory/Outer Measuring Intervals

Outer Measuring Intervals
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 $$\varepsilon\in\Bbb R^+$$ the set $$\{(a-\varepsilon,b+\varepsilon)\}$$ is always an open interval over-approximation.

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

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 $$\mathfrak O$$ 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 $$e'\le e$$.

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 $$\mathfrak O$$.

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 $$e''\le e'$$.

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 $$a_1<a_2<\cdots<a_n$$.

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 $$\mathfrak O''$$. 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 $$e''\ge b-a$$. Reason through the remainder of the proof.

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 $$b-a-2\varepsilon \le \lambda^*((a,b))$$. 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.