MA2224 (Lebesgue integral) Tutorial sheet 3
[February 5, 2016]
Name: Solutions
1. Let I be the (usual) interval algebra (of subsets of R) and define µ : I → [0, ∞] by
(
0 if m(E) < ∞
µ(E) =
1 if m(E) = ∞
[Aside: This will not be a useful way to do things.]
Show that µ is monotone and finitely subadditive, but not finitely additive.
Solution: µ monotone: if E1 , E2 ∈ I with E1 ⊆ E2 , we consider the two cases m(E1 ) <
∞ and m(E1 ) = ∞ separately. If m(E1 ) < ∞, then µ(E1 ) = 0 ≤ µ(E2 ). If m(E1 ) = ∞,
then since m is monotone, ∞ = m(E1 ) ≤ m(E2 ) and so m(E2 ) = ∞ also. Thus
1 = µ(E1 ) ≤ 1 = µ(E2 ) holds in this case also.
µ finitely subadditive: it is enough to show that if E1 , E2 ∈ I , then µ(E1 ∪ E2 ) ≤
µ(E1 ) + µ(E2 ). If m(E1 ∪ E2 ) < ∞, then µ(E1 ∪ E2 ) = 0 ≤ µ(E1 ) + µ(E2 ). If
m(E1 ∪ E2 ) = ∞, then since m is finitely subadditive m(E1 ∪ E2 ) ≤ m(E1 ) + m(E2 )
and so at least one of m(E1 ), m(E2 ) is ∞. So one of µ(E1 ), µ(E2 is 1 and so
1 = µ(E1 ∪ E2 ) ≤ µ(E1 ) + µ(E2 ).
µ not finitely additive: if E1 = (−∞, 0] and E2 = (0, ∞), then E1 , E2 ∈ I are disjoint
and
1 = µ(E1 ∪ E2 ) = µ(R) 6= 2 = µ(E1 ) + µ(E2 ).
2. If S = {x0 } has just one point x0 ∈ R, show that m∗ (S) = 0.
Solution: For ε > S
0, consider E1 = (x0 − ε, x0 ] ∈ I , En = ∅ ∈ I (for n = 2, 3, . . .).
Then S = {x0 } ⊆ ∞
n=1 En and
∞
X
m(En ) = m(E1 ) + 0 = ε.
n=1
Thus m∗ (S) ≤ ε. As ε > 0 is arbitrary, we must have m∗ (S) = 0.
3. If F ⊆ R is a countable set, show that m∗ (F ) = 0.
S
∗
Solution: If F = {x1 , x2 , . . .} is countably infinite, F = ∞
n=1 {xn }. As m is countably
subadditive
∞
X
∗
m (F ) ≤
m∗ ({xn }) = 0
n=1
∗
(using Q2). Hence m (F ) = 0 in this case.
If F = {x1 , x2 , . . . , xn } is finite (countable sets can be finite or countably
S infinite), we can
use finite subadditivity of m∗ in the same way to get m∗ (F ) = 0 (F = nj=1 {xj }).
4. Show carefully that m∗ ([a, b]) = b − a if a ≤ b. [Hint: if a < b compare (a, b] with [a, b].
If a = b, what is [a, b]?]
Solution: As (a, b] ⊂ [a, b] and m∗ is monotone, m∗ ((a, b]) ≤ m∗ ([a, b]). But m∗ ((a, b]) =
m((a, b]) = b − a since (a, b] ∈ I (even if a = b when (a, b] = (a, a] = ∅). So
b − a ≤ m∗ ([a, b]).
By finite subadditivity of m∗ applied to [a, b] = {a} ∪ (a, b] we also have
m∗ ([a, b]) ≤ m∗ ({a}) + m∗ ((a, b]) = 0 + m((a, b]) = b − a
(using Q2).
Richard M. Timoney
2