# Riemann integration in R^n

1. Feb 24, 2010

### snipez90

1. The problem statement, all variables and given/known data
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.

3. 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