# Riemann Integral in Probability With Martingales by Williams

"Note on the Riemann integral" in chapter 5 of Probability with Martingales by Williams reads:

If, for example, f is a non-negative Riemann integrable function on $([0,1],\mathcal{B}[0,1],Leb)$ with Riemann integral I, then there exists an increasing sequence of elements $(L_n)$ of elements of $SF^+$ and a decreasing sequence $(U_n)$ of elements of $SF^+$ such that

$L_n \uparrow L \leq f, U_n \downarrow U \geq f$

and $\mu(L_n) \uparrow I$,$\mu(U_n) \downarrow I$. If we define

$\tilde{f} = \begin{cases} L & \text{ if$L=U$} \\ 0 & \text{otherwise} \\ \end{cases}$

then it is clear that $\tilde{f}$ is Borel measurable, while (since $\mu(L) = \mu(U) = 1$) $\{f \neq \tilde{f}\}$ is a subset of the Borel set $\{L \neq U\}$ which Lemma 5.2b showed to be of measure 0

Here, $SF^+$ is the collection of non-negative simple functions

and $\mu(f) :=: \int_{[0,1]} f(s)\mu(ds)$.

My question is: What are L and U?

• I know $L(P,f)$ as the lower Riemann sum of $f$ using partition $P$
• This would lead me to think that L would be the lower Riemann integral (the supremum of $L(P,f)$ taken over the partitions $P$)

BUT

• The $L_n$ are functions, not numbers, and f is a function as well, so the convergence $L_n \uparrow L$ and inequality $L \leq f$ don't make sense.

So, what are $U$ and $L$?

Thank you very much in advance.

I believe what's meant is that the sequence of functions $L_n$ converges to a function $L$ with the property that $L(x) \le f(x)$ for every $x$, and similarly for the $U_n$.
• That makes sense! Any idea on why $\mu(L) = 1$? Or should that be $\mu(L) = I$? – user2449397 Aug 26 '14 at 15:21
• Almost certainly $\mu(L) = I$, since they've earlier established that $\mu(L_n) \to I$. I think that the "1" is just a typo. – John Hughes Aug 26 '14 at 15:24
Unless otherwise specified, statements about convergence, inequalities, etc, of functions are usually meant to be interpreted pointwise. So "$L_n \uparrow L$" means $L_n(x) \uparrow L(x)$ for every $x$, and "$L \le f$" means $L(x) \le f(x)$ for every $x$.