32.2. SIMPLE FUNCTIONS, σ ALGEBRAS, MEASURABILITY 635

Lemma 32.2.5 Let f : Ω → (−∞,∞] where F is a σ algebra of subsets of Ω. Thefollowing are equivalent.

f−1((d,∞]) ∈ F for all finite d,

f−1((−∞,d)) ∈ F for all finite d,

f−1([d,∞]) ∈ F for all finite d,

f−1((−∞,d]) ∈ F for all finite d,

f−1 ((a,b)) ∈ F for all a < b,−∞ < a < b < ∞.

Definition 32.2.6 Any of these equivalent conditions in the above lemma is what ismeant when we say that f is measurable.

Proof of the lemma: First note that the first and the third are equivalent. To see this,observe

f−1([d,∞]) = ∩∞n=1 f−1((d −1/n,∞]),

and so if the first condition holds, then so does the third.

f−1((d,∞]) = ∪∞n=1 f−1([d +1/n,∞]),

and so if the third condition holds, so does the first.Similarly, the second and fourth conditions are equivalent. Now

f−1((−∞,d]) = ( f−1((d,∞]))C

so the first and fourth conditions are equivalent. Thus the first four conditions are equivalentand if any of them hold, then for −∞ < a < b < ∞,

f−1((a,b)) = f−1((−∞,b))∩ f−1((a,∞]) ∈ F .

Finally, if the last condition holds,

f−1 ([d,∞]) =(∪∞

k=1 f−1 ((−k+d,d)))C ∈ F

and so the third condition holds. Therefore, all five conditions are equivalent.From this, it is easy to verify that pointwise limits of a sequence of measurable functions

are measurable.

Corollary 32.2.7 If fn (ω)→ f (ω) where all functions have values in (−∞,∞], then ifeach fn is measurable, so is f .

Proof: Note the following:

f−1((b+

1l,∞]

)= ∪∞

k=1 ∩n≥k f−1n

((b+

1l,∞]

)⊆ f−1

([b+

1l,∞

])This follows from the definition of the limit. Therefore,

f−1 ((b,∞]) = ∪∞l=1 f−1

((b+

1l,∞]

)= ∪∞

l=1 ∪∞k=1 ∩n≥k f−1

n

((b+

1l,∞]

)⊆ ∪∞

l=1 f−1([

b+1l,∞

])= f−1 ((b,∞])