Is the Riemann integral of a strictly positive function positive?

Here is a "direct" proof that does not require knowledge of the continuity properties of Riemann-integrable functions:

• Suppose by way of contradiction that $\int_0^1 f = 0$.
• It follows that the upper Darboux sums can be made arbitrarily small. Choose some partition $\mathcal{P}_1$ such that $U(f; \mathcal{P}_1) < 1$.
• There must be some subinterval $I_1$ of $\mathcal{P}_1$ on which the supremum of $f$ is less than 1, as otherwise, summing over all subintervals contradicts the condition that $U(f; \mathcal{P}_1)< 1$.
• Choose some partition $\mathcal{P}_2$ for which $U(f; \mathcal{P}_2) < \frac{|I_1|}{2}$. We may assume WLOG that $\mathcal{P}_2$ is a refinement of $\mathcal{P}_1$, as otherwise re replace $\mathcal{P}_2$ by their common refinement.
• Of the subintervals of $I_1$ which are in $\mathcal{P}_2$, their must be one, which we call $I_2$, on which the supremum of $f$ is less than $\frac{1}{2}$. If not, the sum of $M_k \Delta x_k$ over the subintervals of $I_1$ would yield at least $\frac{|I_1|}{2}$, and $U(f; \mathcal{P}_2)$ would be at least this much.
• Similarly, choose $\mathcal{P}_3$ a refinement of $\mathcal{P}_2$ with the property $U(f; \mathcal{P}_3) < \frac{|I_2|}{3}$.
• Continuing in this manner, we obtain a nested sequence $I_1 \supseteq I_2 \supseteq \dots$ of closed bounded intervals with the property that $f(x) < \frac{1}{n}$ for all $x \in I_n$.
• By the Nested Interval Property, the intersection $\bigcap_{n=1}^\infty I_n$ is nonempty. But any point $x$ in this intersection must have the property $\forall n \in \mathbb{N}: f(x) < \frac{1}{n}$, which implies $f(x) = 0$.

If $f$ is non-negative, Riemann integrable and $\int_a^b f(x) dx = 0$ (with $a<b$) then it must be the case that $f(x) = 0$ a.e. Hence if $f$ is strictly positive on $[a,b]$, then it must be the case that $\int_a^b f(x) dx > 0$.

This is straightforward to see using the Lebesgue integral.

See Corollary 3 in www.math.sc.edu/~schep/riemann.pdf for a straightforward proof. The essence is to show that for all $c > 0$ the set $\{x \in [a, b] | f (x) \ge c \}$ has content zero.

Let $f : I \to \mathbb{R}$ be a function on some interval $I$.

If $f$ is continuous and positive on $I$, then $\displaystyle \int_I f >0$. Indeed, $f \geq \alpha >0$ on some closed interval $K \subset I$, so $\displaystyle \int_I f \geq \int_K f \geq \alpha \cdot \mu(K)>0$.