14.14. EXERCISES 349

X−1 (U) ∈ F whenever U is open. Now define a measure on B (R) denoted byλ X and defined by λ X (E) = P({ω : X (ω) ∈ E}) . Explain why this yields a welldefined probability measure on B (R). This is called the distribution measure.

8. Let (Ω,F ) be a measurable space and let f : Ω → R be a measurable functionmeaning that f−1 (U) ∈ F whenever U is open. Then σ ( f ) denotes the small-est σ algebra such that f is measurable with respect to this σ algebra. Show thatσ ( f ) =

{f−1 (E) : E ∈B (R)

}.

9. There is a monumentally important theorem called the Borel Cantelli lemma. It saysthe following. If you have a measure space (Ω,F ,µ) and if {Ei} ⊆ F is suchthat ∑

∞i=1 µ (Ei) < ∞, then there exists a set N of measure 0 (µ (N) = 0) such that if

ω /∈ N, then ω is in only finitely many of the Ei. Hint: You might look at the set ofall ω which are in infinitely many of the Ei. First explain why this set is of the form∩∞

n=1∪k≥n Ek.

10. Let (Ω,F ,µ) be a measure space. A sequence of functions { fn} is said to convergein measure to a measurable function f if and only if for each ε > 0,

limn→∞

µ ({ω : | fn (ω)− f (ω)|> ε}) = 0

Show that if this happens, then there exists a subsequence{

fnk

}and a set of measure

N such that if ω /∈ N, then limnk→∞ fnk (ω) = f (ω) . Also show that if µ is finite andlimn→∞ fn (ω) = f (ω) , then fn converges in measure to f .

11. Prove Chebyshev’s inequality µ ({ω : | f (ω)|> λ})≤ 1λ

∫| f |dµ .

12. ↑Use the above inequality to show that if limn→∞

∫| fn|dµ = 0, then there is a set of

measure zero N and a subsequence, still called { fn} such that for ω /∈ N, fn (ω)→ 0.Hint: Get a subsequence using the above Chebyshev inequality, still denoted as{ fn} , such that µ

([| fn (ω)|> 1

n

])< C2−n Then use the Borel Cantelli lemma of

Problem 9 to conclude that off a set of measure zero, | fn (ω)| < 1n for all n large

enough.

13. Let {rn}∞

n=1 be an enumeration of the rational numbers in [0,1] meaning that everyrational number is included in {rn}∞

n=1 for some n and let fn (x) = 0 except for whenx∈ {r1, · · · ,rn} when it is 1. Explain why fn is Riemann integrable and has Riemannintegral 0. However, limn→∞ fn (x) ≡ f (x) is 1 on rationals and 0 elsewhere so thisisn’t even Riemann integrable with respect to the integrator F (t) = t. It can beshown that the two integrals give the same answer whenever the function is Riemannintegrable. Thus the Lebesgue integral of fn will be 0. So what is the Lebesgueintegral of the function which is 1 on the rationals and 0 on the irrationals? Explainwhy this is so.

14. In fact, µF being only defined on B (R) might not be a complete measure. Thismeans that you can have A⊆ B⊆C and µ (A) = µ (C) with both A,C measurable butB is not. Give a way to enlarge B (R) to a larger σ algebra, extending µ so that theresult is a σ algebra with measure which is a complete measure space, meaning thatif A⊆B and µ (B) = 0 with B measurable, then A is also measurable. Hint: Let a nullset be one which is contained in a measurable set of measure zero. Denoting such setswith N, let F ≡{A∪N where N is a null set, A ∈B (R)} and let µ̂ (A∪N)≡ µ (A).