Riemann integration in R^n

[snipez90]

Homework Statement

a) Let A in R^n be compact, and let f: A -> R be continuous and also non-negative. Show that if there exists some a in A with f(a) > 0, then $\int_{A}f > 0$
b) Now let A in R^n be a closed rectangle, and suppose f: A -> R be bounded and integrable. Show that if f(x) > 0 for all x in A, then $\int_{A}f > 0.$

2. Relevant theorems
A function is Riemann integrable iff its set of discontinuities has measure zero.

The Attempt at a Solution

a) For this, all we need is the fact that continuous functions are sign preserving right (of course f is integrable)? Specifically we can find a $\delta > 0$ such that for all x in the ball of radius $\delta,$ $|f(x) - f(a)| < f(a) \Rightarrow f(x) > 0.$ Then we can choose our partition so that some subrectangle, say S_i is contained in the ball. Hence the supremum (or infimum, I don't think it matters) of f over this subrectangle is actually equal to f(y) for some y in S_i due to compactness, so the upper sum (or lower sum) is > 0 (> 0 contribution from f over S_i, f is nonnegative elsewhere), so we are done. Does this work?

b) I'm less sure about this one, but I think it follows from a) since we can find infinitely many points on A at which f is continuous (due to the relevant theorem), and we really only need one point. Again sups and infs are achieved due to compactness which follows from Heine-Borel.

Thanks

 

[mighty2000]

for your post! Your approach for part a) seems correct. Since f is continuous and non-negative, we can find a small enough ball around a such that f(x) is always positive in this ball. Then, as you mentioned, we can choose a partition that contains this ball and the upper sum will be positive due to the contribution from this subrectangle.

For part b), I think your reasoning is correct. By the relevant theorem, we can find infinitely many points where f is continuous, and therefore, we can find a ball around each of these points such that f(x) is always positive in this ball. Then, we can choose a partition that contains all of these balls and the upper sum will be positive due to the contributions from each subrectangle. Therefore, the overall integral must also be positive.