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Proof: If $\int_a^b f(t) \, dt$ exists and is positive, then there is a subinterval $I$ of $[a,b]$ and a constant $m>0$ such that $f(x) \geq m$ throughout $I$. Hint: Consider $L(f,P)$

user289293
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  • Βy the way, your claim that $f$ is bounded below by a positive constant does not hold, eg $f(x)=1$ on $[0,1]$ except $0.5$ and $f(0.5)=-1$. If $f$ was continuous though you would be right. – Nikolaos Skout Feb 15 '16 at 20:09

2 Answers2

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Since the Riemann integral exists, it is equal to $\sup\{L(f,P): P~ \text{partition of}~ [a,b]\}$, therefore there is a partition $P$ of $[a,b]$ such that $L(f,P)>0$. Therefore there is an interval $I_k$, defined by the above partition, such that $m_k=\inf\{f(x):x \in I_k\}>0$. So $ f(x)\geq m_k>0$ in $I_k.$

Kamil Jarosz
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Nikolaos Skout
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Hint: You can find sum of characteristic functions such that $f_n\leq f$ and $\int_a^bf(t) \, dt = \lim \int_a^bf_n(t)\,dt$. So there exists $n$ such that $\int_a^bf_n \, dt>0$. Write $f_n=\sum a_j\chi_{I_j}$. There exists $j: a_j>0$ thus the restriction of $f$ to $I_j$ is superior to $a_j$.

Michael Hardy
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Tsemo Aristide
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  • Note that one should write $\lim\int_a^b$ rather than $lim\int_a^b$. I changed it. It's coded as \lim\int_a^b. Writing \lim instead of lim not only prevents italicization, but also results in proper spacing in things like $a\lim b$ and $a\lim(b)$ (note that there's less space on the right in $a\lim(b)$ than in $a\lim b$), and when this is displayed rather than inline, it affects the format of the subscripts, as in $\displaystyle a\lim_{n\to\infty} b$. $\qquad$ – Michael Hardy Feb 15 '16 at 19:49
  • Sry, I still don't quite understand it. Could you plz be more specific? – user289293 Feb 15 '16 at 19:54