36 CHAPTER 3. INTEGRALS, FUNCTIONS OF ONE VARIABLE

Now if ∥P∥ ,∥Q∥< δ m,let R = P∪Q. Then∣∣∣∣∣∑P f −∑Q

f

∣∣∣∣∣ ≤∣∣∣∣∣∑P f −∑

Rf

∣∣∣∣∣+∣∣∣∣∣∑R f −∑

Qf

∣∣∣∣∣≤ 1

m(b−a)+

1m(b−a) =

2m(b−a)

Let M ≥max{| f (x)| : x ∈ [a,b]} . Then all Riemann sums are in the interval

[−M (b−a) ,M (b−a)]

Now let

Sn ≡

{∑P

f : ∥P∥< δ n

}Then Sn ⊇ Sn+1 for all n thanks to the fact that the δ n are decreasing. Let

In = [inf(Sn) ,sup(Sn)]

These are nested intervals contained in [−M (b−a) ,M (b−a)] and so there exists I con-tained in them all. However, from the above computation,

sup(Sn)− inf(Sn)≤2n(b−a)

and so there is only one such I. Hence for any ε > 0 given, there exists δ > 0 such that if∥P∥< δ , then ∣∣∣∣∣∑S

f − I

∣∣∣∣∣< ε ■

We say that a bounded function f defined on an interval [a,b] is Riemann integrable ifthe above integral exists. This is written as f ∈ R([a,b]). The above just showed that everycontinuous function is Riemann integrable.

Not all bounded functions are Riemann integrable. For example, let x ∈ [0,1] and

f (x)≡

{1 if x ∈Q0 if x ∈ R\Q

(3.3)

This has no Riemann integral because you can pick a sequence of partitions Pn, such that∥Pn∥ < 1/n and each partition point is rational. Then for your Riemann sums, take thevalue of the function at the left end point. The resulting Riemann sum will always equal 1.But you could just as easily pick your point yi in the Riemann sum to equal an irrationalnumber and these Riemann sums will all equal 0. Therefore, the condition for integrabilityis violated for ε = 1/4.

If you can partition the interval [a,b] into finitely many intervals [zi−1,zi], such that afunction f is continuous on each [zi−1,zi] , then the function will be integrable on [a,b].This is roughly the claim of the next theorem.

Definition 3.0.10 A bounded function f : [a,b]→R is called piecewise continuous if thereare points zi such that a= z0 < z1 < · · ·< zn = b and continuous functions gi : [zi−1,zi]→Rsuch that for t ∈ (zi−1,zi) ,gi (t) = f (t).