Verified answer

Where to start on this question...

Usually, Riemann Integration is done with partitions. However, it looks as though this question is asking for the step function approach. I hope you have discussed these in your course a bit already. Here we go...

Preliminary note: You should probably say that f is bounded on [a, b] as well. (Continuity on [a, b] is a weaker condition than this.) This implies that α <= f(x) <= β for some constants α, β with x in [a, b].

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Note: Since α <= f, the set A = {∫(a to b) φ(x) dx : φ <= f, φ step function} is not empty.

Moreover, α <= f(x) <= β ==> ∫(a to b) φ(x) dx <= β(b - a). Thus, A is bounded above.

So, A has a finite least upper bound (called the "lower Riemann integral", as it's well-defined.)

Similarly, there's the "upper Riemann integral" as well. A function is Riemann integrable iff the upper and lower integrals are equal. Onto the proof.

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(==>) Suppose that f is Riemann integrable on [a, b], and let ε > 0 be given.

By the above note and the definitions of infimum and supremum,

there exist step functions ϕ and ψ such that ϕ <= f <= ψ on [a, b] and

∫(a to b) f(x) dx - ε/2

< ∫(a to b) ϕ(x) dx, via infimum

<= ∫(a to b) f(x) dx, via above triple inequality

<= ∫(a to b) ψ(x) dx

< ∫(a to b) f(x) dx + ε/2, via supremum

Thus, 0 <= ∫(a to b) ϕ(x) dx - ∫(a to b) ψ(x) dx = ∫(a to b) [ϕ(x) - ψ(x)] dx < ε.

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(<==) Now let ε > 0 be given and assume that we have step functions ϕ and ψ with

ϕ <= f <= ψ on [a, b] so that

0 <= ∫(a to b) ϕ(x) dx - ∫(a to b) ψ(x) dx = ∫(a to b) [ϕ(x) - ψ(x)] dx < ε.

However, for any bounded function f on [a, b]:

∫(a to b) ϕ(x) dx

<= [lower Riemann]∫(a to b) f(x) dx

<= [upper Riemann]∫(a to b) f(x) dx

<= ∫(a to b) ψ(x) dx.

Therefore, 0 <= [lower Riemann]∫(a to b) f(x) dx - [upper Riemann]∫(a to b) f(x) dx < ε.

Since ε > 0 is arbitrary, the upper and lower riemann integrals are equal.

Thus, f is Riemann Integrable on [a, b].

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Reference: "Lebesgue Measure and Integration, An Introduction" by Frank Burk, pp. 150-153.

I hope this helps!