Lebesuge Outer Measure of a Set

Question

For a subset E of R(set of real numbers) and k > 0, let kE = {ke | e ∈ E}.

Show that m^*(kE) = k*(m^∗(E)) , where m^*(E) is the Lebesgue outer meausre of the set E.

Answer 1

Here’s a clean, human-readable proof using the definition of Lebesgue outer measure.

Let E ⊂ R and k > 0. Recall

∞ ∞

m^* (A) = inf { ∑ ℓ ( I_n ) : A ⊂ ∪ I_n, I_n intervals },

n = 1 n = 1

where ℓ ([ a, b ]) = b - a (same length for open/half-open variants).

We’ll prove both inequalities and conclude equality.

( ≤ ) Direction

Take any interval cover { I_n } of E. Then { kI_n } (scale each interval by k) covers kE. Lengths scale linearly:

ℓ ( k I_n ) = k ℓ ( I_n ).

Hence

m^* ( k E ) ≤ ∑ ℓ ( k I_{n )} = k ∑ ℓ ( I_n).

n n

Infimizing over all covers { I_n } of E gives

m^* ( k E ) ≤ k m^* ( E ).

≥ Direction

Now take any interval cover { J_n } of k E. Then { ( 1/k ) J_n } covers E, and

ℓ (( 1 / k ) J_n ) = 1/k ℓ ( J_n ).

Thus

m^* ( E ) ≤ ∑ ℓ (( 1/ k ) J_n ) = 1/k ∑ ℓ ( J_n ).

Multiply by k:

k m^* ( E ) ≤ ∑ ℓ (J_n ).

n

Infimizing over all covers { J_n } of kE yields.

k m^* ( E ) ≤ m^* (k E).

Combining,

m^* ( k E ) = k m^* ( E ).

(Edge cases: if m^* ( E ) = ∞ the statement is immediate; for completeness, when k = 0.

we’d have 0 E = { 0 } and m^* ({0}) = 0 = 0 • m^* ( E ), but the problem assumes k > 0. )