Integral of a positive function Let $f$ be Riemann integrable on $[a, b]$, and suppose that $f(x) > 0$ for all
$x \in [a, b]$. Show that $\int\limits_{a}^{b} f(x) dx > 0.$ In the case, $f$ is continuous, we can use the $\delta$ to create a certain partition to get a contradiction. However, that method doesn't work for this problem. Thank you in advance for your help.
 A: If $f$ is Riemann integrable then the set of discontinuities has measure zero.
In particular, we can find some $x \in (a,b)$ such that $f$ is continuous at $x$, and since $f(x) >0$, the result follows. 
A: The following solution doesn't need measure theory, but the Baire category theorem. You have to decide by yourself if this counts as an "elementary" solution.

Since $f > 0$, we can write
$$[a, b] = \bigcup \limits_{n = 1}^\infty \{x \mid f(x) > \tfrac{1}{n}\} \subset \bigcup \limits_{n = 1}^\infty \overline{\{x \mid f(x) > \tfrac{1}{n}\}}$$
Now $[a, b]$ is a complete space that we have written as a union of closed sets. By the Baire category theorem, not all of these sets are nowhere dense.
This means that there is an $N \in \mathbb{N}$, $x_0 \in [a, b]$ and $\varepsilon > 0$ with $U_\varepsilon(x_0) \cap [a, b] \subset \overline{\{x \mid f(x) > \tfrac{1}{N}\}}$. Wlog we may choose $x_0 \in (a, b)$ and $\varepsilon $ in such a way that $U_\varepsilon(x_0) \subset (a, b)$.
From our inclusion we can infer that for every proper interval $[c, d] \subset [a, b]$ with $[c, d] \cap U_\varepsilon(x_0) \ne \emptyset$ there is a $y \in [c, d]$ with $f(y) > \frac{1}{N}$. But this means that every upper Darboux sum of $f$ is bounded from below by the value $\frac{2\varepsilon}{N}$, which implies $\int_a^b f(x) \, dx \ge \frac{2\varepsilon}{N} > 0$.
A: I'm not sure how to do better than the classic proof in general. However, here is an alternative if both $f$ and $1/f$ are integrable.
If $f(x) > 0$ for all $x \in[a,b]$ then it is easy to show that
$$\int_a^b f(x) \, dx \geqslant 0,$$
since for any partition $P= (x_0,x_1, \ldots, x_n)$ and lower sum $L(P,f)$ we have 
$$0 \leqslant \sum_{k=1}^n \inf_{x \in [x_{k-1},x_k]} f(x) (x_k - x_{k-1})= L(P,f),$$
and, hence, 
$$0 \leqslant \sup_PL(P,f) = \int_a^b f(x) \,dx.$$
By the Cauchy-Schwartz inequality
$$0 < (b-a)^2 = \left(\int_a^b \sqrt{f(x)} \sqrt{[f(x)]^{-1}} \, dx\right)^2 \leqslant \int_a^b f(x) \, dx \int_a^b [f(x)]^{-1} \, dx.$$
To avoid a contradiction we must have
$$\int_a^b f(x) \, dx > 0.$$
A: With the almost the same idea of Dominik, I am writing to post the answer without knowledge of measure theory.
We assume by contradiction that $\int\limits_a^b f(x)dx=0$. We need to find the point $x_0$ such that $f(x_0)=0$. 
We first prove: let $g$ be Riemann integrable on $[c,d]$ and $\int\limits_c^d g(x)dx=0$. Then, for any $\epsilon>0$, there is an interval $[c',d']$ on which $g(x)\le \epsilon$. To see this, let $P$ be a partition of $[c, d]$ such that $U(P,g)< \epsilon (d-c)$. Suppose that for each partition interval there is $\xi_i\in [x_{i-1},x_i]$ such that $g(\xi_i)>\epsilon$. Then $U(P,g)> \epsilon (d-c)$ contradicting the choice of $P$. Hence $g(x)\le \epsilon$ on at least one partition interval.
We now apply this to create a sequence of nested intervals $[a_n,b_n]\subset [a_{n-1},b_{n-1}]$ such that $f(x)\le \frac1n$ on $[a_n,b_n]$. Let $x_0$ be a common point of all these intervals (intersection of closed and nested compact sets is nonempty). Then $f(x_0)<\frac1n$ for all $n$, so $f(x_0)=0$.
