If $f, g$ are continuous and $f(x) \ge g(x)$ for all $x \in [a, b]$, then $\int_{a}^{b} f(x) \ dx \ge \int_{a}^{b} g(x) \ dx$. Is the inverse true?

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The proof for the first statement is trivial using the definition of Riemann sums, since

\begin{equation} \int_{a}^{b} f(x) \ dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x \ge \lim_{x \to \infty}\sum_{i=1}^{n}g(x_i^*)\Delta x = \int_{a}^{b} g(x) \ dx \end{equation}

for any sample point $x_i^*$.

But is the inverse true? Namely, if $\int_{a}^{b} f(x) \ dx \ge \int_{a}^{b} g(x) \ dx$, then $f(x) \ge g(x)$ for all $x \in [a, b]$. I couldn't think of any counterexamples where this is true, and trying to adapt the above 'proof' did not work well since we cannot conclude that $f(x) \ge g(x)$ for every $x \in [a,b]$ simply from the sums alone.

Is there a mistake in my thinking? Otherwise, a relevant counter-example (if it is wrong) or a hint for a proof would be greatly appreciated!