MATH20180 Problem Sheet 5

1. Let f : R→ R be a measurable function.

(a) Let B = x ∈ Ω : f(x) > a . We have f ≥ a1B , so by Proposition 41 weget

∫f · dλ ≥

∫a1B · dλ = aλ(B) =∞ .

(b) This is a consequence of the previous question:

If λ(x ∈ R : f(x) > a) =∞ then by the previous question∫f ·dµ =∞ ,

which contradicts that f is Lebesgue-integrable (because here f = f+

and by definition of Lebesgue-integrable we must have∫f+ · dλ finite)..

(c) We take f(x) = 11+x2 as suggested. Then x ∈ R : f(x) > 0 = R does

not have finire Lebesgue measure, but∫

11+x2 ·dλ = [arctan(x)]+∞−∞ is finite,

so f is Lebesgue integrable (we use here that f = f+ and f− = 0 ).

2. To show that f is measurable, we must show that f−1(B) ∈ B(R) for everyB ∈ B(R) (recall that if we do not specify otherwise, we use the σ -field ofBorel sets on R ). We first compute f−1(B) using the definition of f on eachof the Ai :

f−1(B) = x ∈ R : f(x) ∈ B (it is the definition)

=n⋃

i=1

x ∈ Ai : f(x) ∈ B

(because each x ∈ R has to belong to one of the Ai)

=n⋃

i=1

x ∈ Ai : fi(x) ∈ B

(because if x ∈ Ai then f(x) = fi(x))

=n⋃

i=1

Ai ∩ f−1i (B).

Since fi is measurable, f−1i (B) ∈ B(R) and thus Ai ∩ f−1i (B) ∈ B(R) (sinceAi ∈ B(R) ) and by taking the union we obtain f−1(B) ∈ B(R) .

3. A straightforward computation gives∫fn · dλ = 1 (recall that since fn is

Riemann-integrable, the computation is the same as the Riemann integral).Therefore limn(

∫fn · dλ) = 1 .

But, for any x ∈ R we have limn fn(x) = 0 , so limn fn = 0 . Therefore∫(limn fn) · dλ =

∫0 · dλ = 0 . It shows that we cannot always swap integral

and limit. The conditions under which we saw that it can be done are notsatisifed here.

4. (a) By definition of the integral of simple functions we have∫h · dµ =

∫c1A · dµ = cµ(A) = 0.

MATH20180 Problem Sheet 5

(b) Let A = x ∈ Ω : f(x) 6= g(x) = x ∈ Ω : f(x) − g(x) 6= 0 . Byhypothesis we know that µ(A) = 0 . Observe also that outisde of A , wehave f = g , so if there is a difference between

∫f · dµ and

∫g · dµ it

can only come from “inside” A . The idea is that since µ(A) = 0 it willnot count.

In order to see this, we express f in such a way that f − g (and thus A )can appear:

f = g + (f − g) so∫f · dµ =

∫g · dµ +

∫(f − g) · dµ . We show that∫

(f − g) · dµ = 0 :∫(f−g)·dµ =

∫(f−g)+·dµ−

∫(f−g)−·dµ . Clearly we have 0 ≤ (f−g)+ ≤

2M1A and by part (a)∫

(f − g)+ · dµ = 0 . Similarly∫

(f − g)− · dµ = 0and the result follows.

5. We clearly see that for every x ∈ R , fn(x) increases to 1 if x ∈ [0, 1] andfn(x) = 0 otherwise (make a picture, it is enough for us, you do not have toprove it formally, it is first-year stuff). Therefore fn ↑ 1[0, 1] . So by Proposition42 limn(

∫fn · dλ) =

∫(limn fn) · dλ =

∫1[0,1] · dλ = 1 .