Real-valued function with strictly positive integral on every subinterval of [0, 1] Assume we have a Riemann-integrable function $h: [0, 1] \to \mathbb R$ such that
\begin{equation}
\int_a^b h(x)dx > 0\quad \forall a, b: 0 \leq a < b \leq 1.
\end{equation}
Further assume that $f, g: [0, 1] \to \mathbb R$ are continuous functions with
\begin{equation}
f(x) \geq g(x) \quad\forall x \in [0, 1].
\end{equation}
I suspect that in this case

\begin{equation}
\int_0^1 f(x)h(x)dx \geq \int_0^1 g(x)h(x)dx.
\end{equation}

However, I cannot prove this (I need this as part of another proof). I would like to stay within "elementary" analysis, i.e., do not resort to measure theory (no "almost everywhere" stuff).
Many thanks.

I have found a proof that works even under weaker conditions. $f$ and $g$ need to be merely Riemann-integrable and $\int_a^b h(x) \, dx \ge 0$ needs to hold (i.e. the integral does not have to be strictly positive).
We can assume wlog $g \equiv 0$.
Let $Z_n$ be a sequence of meshes of $[0, 1]$ with $|Z_n| \to 0$. Since $h$ is Riemann-integrable, we know $\int_0^1 h(x) \, dx = \lim \limits_{n \to \infty} \overline{S}(Z_n, h)$, where $\overline{S}(Z, h) = \sum \limits_{I \in Z} |I| \sup \limits_{x \in I} h(x)$.
Since the integral of $h$ is nonnegative on every interval, we know that $\sup \limits_{x \in I} h(x) \ge 0$ holds for all (nondegenerate) intervals $I \subset [0, 1]$. But from this we can conclude that $\sup \limits_{x \in I} \{f(x) h(x)\} \ge 0$. Since $f \cdot h$ is Riemann-integrable, this yields the assertion:
$$\int f(x) h(x) \, dx = \lim \limits_{n \to \infty} \overline{S}(Z_n, f \cdot h) = \lim \limits_{n \to \infty} \sum \limits_{I \in Z_n} |I| \sup \limits_{x \in I}\{ f(x) h(x)\}  \ge \lim \limits_{n \to \infty} \sum \limits_{I \in Z_n} 0 = 0$$
