# The Integral

What we introduced last time:

$L(f)=\sup _{P}L(P,f)$

$U(f)=\inf _{P}U(P,f)$

Are actually known as the lower and upper integrals respectively. Both are simultaneously the integral provided that the function that they are built from satisfy the following condition, the definition of integrability (but not the integrability criterion).

A bounded function $f$  on $[a,b]$  is integrable if $L(f)=U(f)$ .

And that's all the integral really is!

## Yet ANOTHER return of that summing exercise

This is just a quick little question, probably trivial. What does this mean for $f(x)=x$  ?

# Integrability

As previously defined we can prove the integrability of a function by noting that $U(f)=L(f)$

However, there is a much more useful way to prove that a function, or an entire class of functions, is integrable. This is the theorem called the Integrability Criterion:

A bounded function $f:[a,b]\to \mathbb {R}$  is integrable on $[a,b]$  iff for all $\varepsilon >0$  there exists a partition $P$  of $[a,b]$  such that $U(P,f)-L(P,f)<\varepsilon$

You'll notice that it has properties similar to the definition of the limit.

Since this is a proof of necessity assume that $f$  is integrable on $[a,b]$  . Let $\varepsilon >0$  be given and say $I=\int \limits _{a}^{b}f(x)dx$  . Also $I=\inf _{Q}U(Q,f)=\sup _{R}L(R,f)$  . Now by the approximation property there exists a partition $Q$  such that $U(Q,f)  and a partition $R$  such that $L(R,f)>I-{\tfrac {\varepsilon }{2}}$  . If we take the refinement $P=Q\sup R$  and apply the properties of refinement that we proved before:

$U(P,f)

and

$L(P,f)>I-{\frac {\varepsilon }{2}}$

We can manipulate this to:

$U(P,f)-L(P,f)<\varepsilon$

Now for the proof to finish up the biconditional...

## Integrability criterion proof

You do it. If for all $\varepsilon >0$  there exists a partition $P$  of $[a,b]$  such that $U(P,f)-L(P,f)<\varepsilon$  then $f$  is integrable on $[a,b]$  .

Some hints: show that it implies $U(f)-L(f)<\varepsilon$  for all $\varepsilon >0$  then show this is so only if $U(f)=L(f)$  .