14.5. MEASURABLE FUNCTIONS 331

Definition 14.5.3 Any of these equivalent conditions in the above lemma is what ismeant the statement 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 14.5.4 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,∞])

The messy term on the middle is measurable because it consists of countable unions and in-tersections of measurable sets. It equals f−1 ((b,∞]) and so this last set is also measurable.By Lemma 14.5.2, f is measurable.

A convenient way to check measurability is in terms of limits of simple functions.

Definition 14.5.5 Let (Ω,F ) be a measurable space. A simple function is onewhich is measurable but has only finitely many values.