If $f(x)$ is a real continuous function that is zero almost everywhere on the interval $[a,b]$, how can I prove that it is zero everywhere in that interval?
My apologies for the probably silly question.
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3If it is nonzero on $[a,b]$ then by continuity it is nonzero on an open interval contained in $[a,b]$. – Angina Seng Jun 22 '17 at 14:28
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Thank you @LordSharktheUnknown – user52227 Jun 22 '17 at 14:34
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As stated, I would not accept the conclusion. Perhaps there is some formal definition of "almost everywhere" but as it stands, the premise, as I understand it, doesn't lead to the conclusion. This sounds like the same line of thinking used to justify some conclusions in variational calculus, but the statements there are a bit more precise. Perhaps the reason the question is difficult to answer is that it is poorly stated. – Steven Thomas Hatton Jun 22 '17 at 14:51
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4@StevenHatton I don't think there is any ambiguity here. "Almost everywhere" means everywhere outside a set of Lebesgue measure zero. The question is well-formed and the hypothesis does indeed lead to the conclusion. – User8128 Jun 22 '17 at 14:55
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I just checked my real analysis text book, and it never mentions "Lebesgue measure zero". Without further qualification "almost everywhere" means almost everywhere. – Steven Thomas Hatton Jun 22 '17 at 15:09
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@StevenHatton It is a very standard term mostly introduced in measure theory. There is no clarification needed in what the question asks. – Sahiba Arora Jun 22 '17 at 16:47
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Suppose that $f(x_0) \ne 0$ for some $x_0 \in [a,b]$ . We can asume that $f(x_0) > 0$. Then there is an intervall $J$ in $[a,b]$ with $x_0 \in J$ and $f(x) > 0$ for all $x \in J$. But this is a contradiction to $f(x)=0$ a.e. on $[a,b]$.
Fred
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Hint: suppose not at some point, use continuity to form an interval and then integrate over this interval to derive the contradiction.
operatorerror
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