Let R a1 b1 x x an bn C and let f R rightarrow R be a cont

Let R = [a_1, b_1] x ... x [a_n, b_n] C and let f: R rightarrow R be a continuous function. Prove that f is Riemann integrable on R. Recall that remans integrable on an n-rectangle means that the upper and lower integrals of f agree. (As in class, not in Rudin; he defines this only as an iterated integral, which is a theorem rather than definition for us.) Continuous on a compact set implies uniformly continuous. Refer to the proof of this fact in the single variable case in Chapter 6 of Rudin.

Solution

Solution: To prove the result first we prove the following result:

A continuous function f : [a, b] R on a compact interval is Riemann integrable.

Proof: A continuous function on a compact set is bounded, so we just need to
verify the Cauchy condition.
Let \\epsilon > 0. A continuous function on a compact set is uniformly continuous, so
there exists > 0 such that
|f(x) f(y)| < \\epsilon/(b a) for all x, y [a, b] such that |x y| < .

Choose a partition P = {I₁, I₂, . . . , I_n} of [a, b] such that |I_k| < for every k; for
instance, we can take n intervals of equal length (b a)/n with n > (b a)/.

Since f is continuous, it attains its maximum and minimum values M_k and
m_k on the compact interval I_k at points x_k and y_k in I_k. These points satisfy
|x_k y_k| < , so
M_k m_k = f(x_k) f(y_k) < \\epsilon/(b a).

Theerefore upper and lower sums of f satisfy
U(f; P) L(f; P) = \\sigma_{k=1}^{n}M_k|I_k| \\sigma_{k=1}^{n} m_k|I_k|

= \\sigma_{k=1}^{n}(M_k m_k)|I_k|
= \\epsilon/(b a) \\sigma_{k=1}^{n}|I_k| < \\epsilon

and by The Cauchy criterion for integrability, we have f is integrable in [a,b].

Since [a₁,b₁], [a₂,b₂], ..., [a_n, b_n] all are compact, so their cartesian product is also cmpact.

Hence R = [a₁,b₁]X [a₂,b₂]X ... X [a_n, b_n] is compact.

Therefore by above result , we can say that f is Riemann integrable on R, (Proved}