Then for functions which have values in (−∞,∞] we have the following Lemma which includes a definition of what it means for a function to be measurable.
Notation 5.1.1 In whatever context, f^{−1}
Lemma 5.1.2 Let f : Ω → (−∞,∞] where ℱ is a σ algebra of subsets of Ω. The following are equivalent.





Any of these equivalent conditions is equivalent to the function being measurable.
Proof: First note that the first and the third are equivalent. To see this, observe

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

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

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

Finally, if the last condition holds,

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.
Proof: Note the following:

This follows from the definition of the limit. Therefore,
Observation 5.1.4 If f : Ω → ℝ then the above definition of measurability holds with no change. In this case, f never achieves the value ∞. This is actually the case of most interest.
The following theorem is of major significance. I will use this whenever it is convenient.
Theorem 5.1.5 Suppose f : ℝ → ℝ is measurable and g : ℝ → ℝ is continuous. Then g ∘ f is measurable. Also, if f,g are measurable real valued functions, then their sum is also measurable and real valued as are all linear combinations of these functions.
Proof:
Why is f + g measurable when f,g are real valued measurable functions? This is a little trickier. Let the rational numbers be

It is clear that the expression on the right is contained in
Although the above will usually be enough in this book, it is useful to have a theorem about measurability of a continuous combination of measurable functions. First note the following.
Lemma 5.1.6 Let
Proof: See Problem 2 on Page 473.
Proposition 5.1.7 Let f_{i} : Ω → ℝ be measurable,
Proof: From the above lemma, it suffices to verify that

Since g is continuous, it follows from Proposition 2.2.13 that g^{−1}

Thus

Note that this includes all of Theorem 5.1.5 as a special case.
There is a fundamental theorem about the relationship of simple functions to measurable functions given in the next theorem.
Definition 5.1.8 Let E ∈ℱ for ℱ a σ algebra. Then

This is called the indicator function of the set E. Let s :

where E_{i} ∈ℱ and c_{i} ∈ ℝ, the E_{i} being disjoint. Thus simple functions have finitely many values and are measurable. In the next theorem, it will also be assumed that each c_{i} ≥ 0.
Each simple function is measurable. This is easily seen as follows. First of all, you can assume the c_{i} are distinct because if not, you could just replace those E_{i} which correspond to a single value with their union. Then if you have any open interval

and this is measurable because it is the finite union of measurable sets.
Theorem 5.1.9 Let f ≥ 0 be measurable. Then there exists a sequence of nonnegative simple functions {s_{n}} satisfying
 (5.1) 

 (5.2) 
If f is bounded, the convergence is actually uniform. Conversely, if f is nonnegative and is the pointwise limit of such simple functions, then f is measurable.
Proof: Letting I ≡

Then t_{n}(ω) ≤ f(ω) for all ω and lim_{n→∞}t_{n}(ω) = f(ω) for all ω. This is because t_{n}
 (5.3) 
Thus whenever ω

Then the sequence {s_{n}} satisfies 5.15.2. Also each s_{n} has finitely many values and is measurable. To see this, note that

To verify the last claim, note that in this case the term 2^{n}X_{I}(ω) is not present and for n large enough, 2^{n}∕n is larger than all values of f. Therefore, for all n large enough, 5.3 holds for all ω. Thus the convergence is uniform.
Now consider the converse assertion. Why is f measurable if it is the pointwise limit of an increasing sequence simple functions?

because ω ∈ f^{−1}