Question 11.T.6: Let f be a bounded function on the compact interval I = [a, ...

Let P = \left\{x_{0}, x_{1}, · · · , x_{n}\right\} be a partition of I, and define the functions \varphi_{P} and ψ_{P} on I as follows:

\varphi_{P} = \sum\limits_{i=0}^{n−1}{m_{i}χ_{[x_{i},x_{i+1})}},

ψ_{P} =\sum\limits_{i=0}^{n−1}{M_{i}χ_{[x_{i} ,x_{i+1})}},

where, as usual,

m_{i} = \inf \left\{f (x) : x_{i} ≤ x ≤ x_{i+1}\right\},

M_{i} = \sup \left\{f (x) : x_{i} ≤ x ≤ x_{i+1}\right\}.

Note that

\varphi_{P} ≤ f ≤ ψ_{P} ,  \varphi_{P} , ψ_{P} ∈ S(I).

Referring to Definitions 8.2 and 8.3,

\int_{I} \varphi_{P} dm = L(f, P ),

\int_{I} ψ_{P} dm = U(f, P ).

Furthermore,

Q ⊇ P ⇒  \varphi_{Q} ≥ \varphi_{P} ,  ψ_{Q} ≤ ψ_{P} .

Choose a sequence of partitions (P_{n}) of I such that, for each n ∈ \mathbb{N},

P_{n+1} ⊇ P_{n},  \left\|P_{n}\right\| → 0.

For example, P_{n} could be the uniform partition of I into 2^{n} equal parts.

Setting \varphi_{n} = \varphi_{P_{n}} and ψ_{n} = ψ_{P_{n}} , we conclude that the sequence (\varphi_{n}) is increasing while (ψ_{n}) is decreasing. Their limits

\varphi(x) = \underset{n→∞}{\lim} \varphi_{n}(x),  ψ(x) = \underset{n→∞}{\lim}  ψ_{n}(x)

are clearly measurable and satisfy φ ≤ f ≤ ψ. Since f is bounded, the sequence (ψ_{n} − \varphi_{n}) is also bounded, and the bounded convergence theorem (Corollary 11.5.1) implies

\int_{I}{(ψ − \varphi)} dm = \underset{n→∞}{\lim} \int_{I} {(ψ_{n} − \varphi_{n})} dm

= \underset{n→∞}{\lim} [U (f, P_{n}) − L(f, P_{n})]

= U(f ) − L(f) = 0,

where we use Darboux’s theorem and the fact that f is Riemann integrable in the last two equalities. Since ψ − φ ≥ 0 on I, it follows from Theorem 11.2(vii) that ψ − φ = 0 a.e., hence

f = φ = ψ a.e.,        (11.24)

which means f is measurable. Furthermore,

\int_{I} {f} dm = \int_{I} {\varphi} dm

= \underset{n→∞}{\lim} \int_{I} {\varphi} dm

= \underset{n→∞}{\lim} L(f, P_{n})

= L(f )

= \int_{a}^{b} {f (x)} dx.