Lebesgue Outer Measure and Length of an Interval

Question

Question:

For an interval of the form (-infinity, b) or (-infinity, b], show that m^*(I) = l(I), where m^*(I) and l(I) denote the Lebesgue Outer Measure of I and the length of interval I, respectively.

My approch towards the question:

Let I be an interval of the form (-infinity, b) or (-infinity, b]

Let k be any positive number.

Then, [b-2k, b-k] is a subset of I.

Then, m^*([b-2k, b-k]) <= m^*(I)

===> (b-k) - (b-2k) <= m^*(I)

===> k<= m^*(I)

Since this is true for any positive number k, we must have,

m^*(I) = infinity = l(I)

My doubts are whether,

(i) k should be any positive number or k should be any positive real number?

(ii) Is this a concrete proof for the required result?

Your help is highly appreciated!

Answer 1

Short answer: yes, your idea works, with one small tweak.

1. Take any positive real k > 0 (not just integer.)
2. Note that [b−2k, b−k]⊂(−∞,b) and also [ b-2k, b-k ] ⊂ ( - ∞, b - k]
3. For bounded intervals we know m^∗([x, y])=y−x (this is a standard lemma proved from the definition of outer measure). Hence

k = m∗ ([b−2k, b−k]) ≤ m∗(i).

1. Since this holds for every k > 0, m^ * (I) cannot be finite; thus m^∗ (I) =∞. By convention the “length” ∫ ( i ) of an unbounded interval is also ∞, so m^∗( i ) = ∫ ( i ).

So:

(i) k should be any positive real number.

(ii) Your proof is correct provided you’re allowed to use the fact that m^* of a bounded interval equals its length. If you want a variant that avoids invoking that directly, do this:

Let J N= ⋃n =1N [ b−n−1, b−n ]⊂ i. For a finite disjoint union of intervals,

m^∗(JN)=∑n=1 1=N. By monotonicity, m^∗(i) ≥ m^∗ (JN)=N m^*(I) for all N, hence m^∗(i) =∞.