MA2224 (Lebesgue integral) Tutorial sheet 5
[February 19, 2016]
Name: Solutions
1. Show that the Borel σ-algebra is the σ-algebra generated by the intervals [a, b) with a < b
(a and b finite).
Solution: Let’s denote the σ-algebra generated by {[a, b) : a < b} by Σ0 and the Borel
σ-algebra by ΣBorel .
By the theorem giving different collections of sets that generate the ΣBorel , we know that
single point sets {a} = [a, a] and open intervals are in ΣBorel . So the ΣBorel must contain
[a, b) = {a} ∪ (a, b)
whenever a < b. So Σ0 ⊆ ΣBorel .
To show that ΣBorel ⊆ Σ0 , it is enough to show Σ0 contains (say) all finite closed intervals
[a, b] (with a ≤ b). What we really need is to show that one point sets are in Σ0 .
But for a ∈ R, we can see that
{a} =
∞ \
n=1
1
a, a +
n
∈ Σ0 .
So [a, a] ∈ Σ0 and for a < b we also have
[a, b] = [a, b) ∪ {b} ∈ Σ0 .
As we know from the theorem that ΣBorel is the smallest σ-algebra that contains all [a, b],
we can conclude that ΣBorel ⊆ Σ0 .
We already had Σ0 ⊆ ΣBorel and so Σ0 = ΣBorel .
2. If f1 , f2 , . . . is a sequence of Lebesgue measurable functions fn : R → R such that
f1 (x) ≤ f2 (x) ≤ f3 (x) ≤ · · ·
holds for all x ∈ R and such that
f (x) = lim fn (x)
n→∞
exists for all x ∈ R, show that f must be a Lebesgue measurable function. [Hint: Because
the sequence of functions is pointwise increasing, f (x) ≤ a holds if and only if fn (x) ≤ a
for all n.]
Solution: Our objective is to show that
f −1 ((−∞, a]) = {x ∈ R : f (x) ≤ a} ∈ L
for each a ∈ R.
Fix a ∈ R.
Usng the hint
{x ∈ R : f (x) ≤ a} =
∞
\
{x ∈ R : fn (x) ≤ a}
n=1
Since fn is Lebesgue measurable,
{x ∈ R : fn (x) ≤ a} ∈ L
T
and so then is the (countable) intersection ∞
n=1 {x ∈ R : fn (x) ≤ a} ∈ L (since σ algebras are
closed under taking countable intersections).
Richard M. Timoney
2