# f > 0 implies int f(x) dx > 0

By | Mai 7, 2015

If $$f : \mathbb{R} \to \mathbb{R}$$ is continuous on $$[a,~b]$$ and $$f > 0$$ then $\int_{a}^{b} f(x) ~ dx > 0 .$ This theorem is a well-known example in undergraduate analysis and can be proven easily. But what if the continuity of $$f$$ is replaced with integrability?

Theorem. If $$f : \mathbb{R} \to \mathbb{R}$$ is Riemann-integrable on $$[a,~b]$$ and $$f > 0$$ on $$[a,~b]$$ then $\int_{a}^{b} f(x) ~ dx > 0 . \tag{1}$

Proof. Obviously $$L(f,~P) \ge 0$$ for any partition $$P$$ of $$[a,~b]$$, thus we have $\int_{a}^{b} f(x) ~dx \ge 0 . \tag{2}$ Now, suppose that $\int_{a}^{b} f(x) ~ dx =0 . \tag{3}$ For $$f$$ is Riemann-integrable, $$f$$ is Lebesgue-integrable and measurable. Let $$M$$ be the supremum of $$f$$ on $$[a,~b].$$ Then the inverse image $f^{-1} ((0,~ M+1)) \tag{4}$ is also measurable and the measure of (4) equals zero by (3). But $0 < f(x) < M+1$ for all $$x \in [a,~b]$$, which implies that the measure of (4) has to be same to the measure of $$[a,~b] .$$ This is a contradiction, i.e., (3) is false; Consequently by (2), we have $\int_{a}^{b} f(x) ~dx > 0 .$