Let $f:[a,b]\to\mathbb{R}$ be a function Riemann-integrable such that $f(x)\geq 0$ for all $x\in[a,b]$.
If $\int_{a}^{b}f=0$ then $\{x\in[a,b]:f(x)=0\}$ is dense set.
Can someone help me with this ?
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2A nice question. It is easy for Lebesgue integral, but how can we do it without using any Lebesgue theory? – GEdgar May 25 '16 at 16:03
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@GEdgar The posted answer, plus your invaluable addition to it, don't use any Lebesgue theory, right? – layman May 25 '16 at 16:06
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So far there is no posted answer, only a hint. How can you complete it without Lebesgue theory? – GEdgar May 25 '16 at 16:08
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@GEdgar See the amazing answer here – MathematicsStudent1122 May 25 '16 at 16:09
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So this question may be considered a duplicate of 1794822... – GEdgar May 25 '16 at 16:18
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Here is a hint: Suppose $A = \{x \in [a,b] : f(x) = 0\}$ is not dense. Then there is some pocket $(c,d)$ in the interval $[a,b]$ untouched by $A$, i.e., $f(x) \neq 0$ for all $x \in (c,d)$. (Why? What does density even mean?)
Then since $f(x) \geq 0$ by assumption, and thus $f(x) > 0$ on $(c,d)$ (since it's not equal to $0$ at any point in this interval), what can you say about $\int \limits_{c}^{d} f(x) \,dx$? What does this imply about $\int \limits_{a}^{b} f(x) \,dx$?
layman
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1So we need to ask: could $f$ be positive everywhere, but have infimum $0$ on every open subinterval? That is, every lower sum is zero. Then how about upper sums? – GEdgar May 25 '16 at 16:00
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Note: a function can, indeed, be everywhere positive but have infimum zero on every open subinterval. Thus every lower sum could be zero. We need to show this is not possible for a Riemann-Integrable function. Easy with Lebesgue theory. To do this without Lebesgue theory, see the "amazing answer" cited by MathematicsStudent. – GEdgar May 25 '16 at 16:16