MA2224 (Lebesgue integral) Tutorial sheet 6
[February 26, 2016]
Name: Solutions
1. If f : R → R is Lebesgue measurable and g : R → R is continuous, show that g ◦ f is
Lebesgue measurable. [Hint: If E ⊆ R is closed then g −1 (E) is a closed set, hence a
Borel set. Also (g ◦ f )−1 (E) = f −1 (g −1 (E)). What do we know about f −1 (B) for B
Borel?]
Solution: The aim is to show that for each a ∈ R we have (g ◦ f )−1 ((−∞, a]) ∈ L .
Taking E = (−∞, a] in the hint, we have
(g ◦ f )−1 ((−∞, a]) = f −1 (g −1 ((−∞, a]))
Since g is continuous and (−∞, a] ⊂ R is closed, we know that g −1 ((−∞, a]) is closed
and hence Borel. (Recall that one way to characterise the Borel σ-algebra is that it is the
σ-algebra generated by the closed sets.)
Recall now the result that if f is Lebesgue measurable then f −1 (B) ∈ L holds for each
Borel set B. So
(g ◦ f )−1 ((−∞, a]) = f −1 (g −1 ((−∞, a])) ∈ L
This is true for each a ∈ R and so f is Lebesgue measurable.
2. Give an example of a function f : R → R that is not Lebesgue measurable. [Hint: is there
a non-measurable set?]
Solution: We know there is a nonmeasurable subset of R, that is E ⊆ R with E ∈
/ L.
Then f = χE is a nonmeasurable function since f −1 ((−∞, 0]) = E c = R \ E ∈
/ L.
3. Let f : R → R be given by f = 3χ[0,2] + 5χ(1/2,1] . Show that f is a simple (Lebesgue)
measurable function and find a standard representation for f .
Solution: Since each characteristic function has only 2 possible values 0 and 1, f can have
at most 4 values 3(0) + 5(0) = 0, 3(1) + 5(0) = 3, 3(0) + 5(1) = 5 and 3(1) + 5(1) = 8.
So f is simple.
Since the inervals [0, 2] and (1/2, 1] are measurable sets, their characteristic functions must
be measurable and so the linear combination f must also be (Lebesgue) measurable.
We have f (x) = 0 when x ∈ R \ ([0, 2] ∪ (1/2, 1]) = R \ [0, 2] = (−∞, 0) ∪ (2, ∞),
f (x) = 8 when x ∈ (1/2, 1], f (x) = 3 when x ∈ [0, 2] \ (1/2, 1] = [0, 1.2] ∪ (1, 2] and
f (x) = 5 never. So f (R) = {0, 8, 3} and the standard form for f is
f = 0χ(−∞,0)∪(2,∞) + 8χ(1/2,1] + 3χ[0,1.2]∪(1,2]
Richard M. Timoney