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 \[\int_{a}^{b} f(x) ~dx > 0 .\]

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