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This present question is inspired by this earlier question.

Consider the problem of proving that if $f\ge 0$ on an interval $I\subseteq\mathbb R$ and $\int_I f=0$, then $f$ is $0$ (almost) everywhere on $I$.

Can it be proved that the only way to prove this is by contradiction?

PS: How about if we consider two cases: (1) $f$ is continuous (and then disregard the word "almost"); (2) $f$ is measurable.

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Michael Hardy
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  • @GitGud : Further edited accordingly. – Michael Hardy Jul 27 '13 at 18:03
  • It's usually proved by contraposition, no? – David Mitra Jul 27 '13 at 18:06
  • I don't think this question is answerable without first specifying precisely what axiomatic system we're working in, and which deduction rules in the underlying logic are deemed to be proofs-by-contradiction. – Daniel Franke Jul 27 '13 at 18:07
  • In the first case, the point of proving it by contradiction is to find a point $a\in I$ such that $f(a)\neq 0$ and then proceed using continuity.

    This can be avoided by considering the logical truth: $$\text{there exists }a\in I \text{ such that }f(a)\neq 0\lor f \text{ is the null function}.$$

    This of course assumes that the above statement is in fact a logical truth, but that's dependent on the law of excluded middle which is equivalent to the principle of noncontradiction.

     – Git Gud Jul 27 '13 at 18:09
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    @GitGud Constructivists would be very surprised to hear that the law of excluded middle (which they reject) is equivalent to the principle of noncontradiction (which they accept). The former says that at least one of the two propositions $p$, $\neg p$ is true; the latter says that at most one of those two propositions is true. – Andreas Blass Jul 27 '13 at 18:17
  • @AndreasBlass I wasn't sure about what I was saying, so I googled it, misread something and ultimately made the mistake you pointed out. Thanks. – Git Gud Jul 27 '13 at 18:19
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    What about this: Define $F(x)=\int_0^x f(t), dt$. Since $f\ge 0$, we have $0\le F(x)\le\int_0^1 f(t),dt=0$ for all $x\in[0,1]$. Thus $F$ is identically $0$ on $[0,1]$. Since $f$ is continuous, we have $F'(x)=f(x)=0$ for all $x\in[0,1]$. – David Mitra Jul 27 '13 at 18:30
  • @AndreasBlass By the way, above I meant double negation instead of LOM. – Git Gud Jul 28 '13 at 02:02
  • Lebesgue Differentiation for question (2)? – user71352 Jul 28 '13 at 02:23
  • @GitGud The law of double negation, specifically the half of it that says "if not not p then p," is. like LEM, rejected in constructive logic, and is in fact constructively equivalent to LEM. The other half of double negation ("if p then not not p") and the law of non-contradiction ("not both p and not p") are constructively valid. – Andreas Blass Jul 28 '13 at 07:17
  • @AndreasBlass I'm sorry for making so many mistakes, what I meant was DN instead of PofnonC. I'm actualy thinking right, just writing the wrong things. I meant DN instead of PofnonC from the start, which is the relevant thing to mention because the OP is looking for a proof without contradiction. – Git Gud Jul 28 '13 at 10:23

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Without contradiction:

For every $n\geqslant1$, let $B_n=\{x\in I\mid nf(x)\geqslant1\}$, then $0\leqslant\mathbf 1_{B_n}\leqslant nf$ (this step uses the hypothesis that $f\geqslant0$ everywhere on $I$) hence $0\leqslant\mathrm{Leb}(B_n)\leqslant n\int\limits_If=0$, which implies that $\mathrm{Leb}(B_n)=0$. Now, the set $B=\{x\in I\mid f(x)\ne0\}$ is $B=\bigcup\limits_{n\geqslant1}B_n$ hence $\mathrm{Leb}(B)\leqslant\sum\limits_{n\geqslant1}\mathrm{Leb}(B_n)=0$, QED.

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