Question 11.T.7: Let f be a bounded function on a compact interval I = [a, b]...
Let f be a bounded function on a compact interval I = [a, b], and suppose C ⊆ I is it domain of continuity. Then
f ∈ \mathcal{R}(a, b) ⇔ m(I \setminus C) = 0,
that is, f is Riemann integrable on [a, b] if, and only if, f is continuous almost everywhere.
Let P_{n}, \varphi_{n}, ψ_{n}, \varphi, ψ be as defined in the proof of Theorem 11.6.
(i) Suppose f ∈ \mathcal{R}(a, b). It suffices to show that
I\C ⊆ E ∪ {x ∈ I : φ(x) ≠ ψ(x)}, (11.25)
where E = \cup_{n=1}^{∞} P_{n} is the set of all points of the partitions P_{n}. That is because, on the one hand, E is countable (hence m(E) = 0) and, on the other, m({x ∈ I : φ(x) ≠ ψ(x)}) = 0 as we saw in equation (11.24).
f = φ = ψ a.e., (11.24)
To prove (11.25), let x ∈ I\C. Then there is a positive number ε and a sequence (x_{k}) in I such that x_{k} → x and |f (x_{k}) − f (x)| ≥ ε for all k.
If x ∉ E then, for every n ∈ \mathbb{N}, there is a k such that x_{k} and x both lie in the same subinterval of P_{n}. But since |f (x_{k}) − f (x)| ≥ ε, we obtain
ψ_{n}(x) − \varphi_{n}(x) ≥ ε
⇒ ψ(x) − φ(x) ≥ ε
⇒ x ∈ {x ∈ I : φ(x) ≠ ψ(x)}.
(ii) Assume now that m(I\C) = 0. If x ∈ C then, for every ε > 0, there is a δ > 0 such that
x^{\prime} ∈ (x − δ, x + δ) ⇒ \left|f (x) − f (x^{\prime})\right| < ε. (11.26)
Since \left\|P_{n}\right\| → 0 we can choose N so that the subinterval of P_{N} which contains x is contained in (x − δ, x+δ). In view of (11.26) we therefore have
ψ_{N} (x) − \varphi_{N} (x) ≤ 2ε
⇒ ψ(x) − φ(x) ≤ 2ε.
Since ε > 0 is arbitrary, ψ(x) = φ(x), and since m(I\C) = 0, ψ = φ a.e. Referring back to equation (11.23), this implies that U(f) = L(f ) and hence f ∈ \mathcal{R}(a, b).
\underset{t→c}{\lim} \int_{Ω}{f_{t}} dm = \int_{Ω}{f} dm. (11.23)