Measure Theory/Outer Measuring Intervals
Outer Measuring Intervals
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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]
[edit | edit source]Now we show that for any closed, bounded interval, . As is typical with proofs in analysis, we do this by showing two inequalities.
The first inequality that we will show is .
We can accomplish this by observing that for any the set is always an open interval over-approximation.
The corresponding over-estimate is .
Exercise 2. Carry the
Complete the argument that, due to what we have seen above, . |
In order to show , we will show that is a lower bound on the set of over-estimates. Recall that, by definition, is the greatest of all such lower bounds.
To this end we let be any open interval over-approximation of . Let be the corresponding over-estimate.
Because is compact, and because is an open cover, then there must exist a finite subcover, . Let be its corresponding over-estimate.
Notice that every term which makes up e' also occurs in e. Therefore .
Further, it will be useful if no interval in is a subset of any other interval in it. Therefore we may construct a still smaller approximation, which results from removing intervals from until no interval that remains is a subset of any other interval in .
Exercise 3. Removing Subset Intervals Is Fine
Prove that constructed above is still an open interval over-approximation of . |
Let be the over-estimate corresponding to . As before, we have .
Now assume that and assume that these intervals are listed "in order". Here "in order" means that .
Exercise 4. Check the Interval Order
1. Prove that, with the ordering given above for the intervals, for each . Hint: If this were not true, you would get an interval inside another interval in . Further show that and . 2. Show that . Hint: Write out the definition of 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 . 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 and then use monotonicity to prove . Then let . The reverse inequality should be even more direct. The use of monotonicity and the outer measure of closed intervals, gives an easy proof that if I is an unbounded interval. |