Here is Prob. 2, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f \geq 0$, $f$ is continuous on $[a, b]$, and $\int_a^b f(x) \ \mathrm{d} x = 0$. Prove that $f(x) = 0$ for all $x \in [a, b]$.

My Attempt:

Let $c$ and $d$ be any points in $[a, b]$ such that $c \leq d$. As $f(x) \geq 0$ for all $x \in [a, b]$ and as $f$ is continuous on $[a, b]$, so $f(x) \geq 0$ for all $x \in [c, d]$ and $f$ is continuous on $[c, d]$; therefore $f \in \mathscr{R}$ on $[c, d]$ and $$ \int_c^d f(x) \ \mathrm{d} x \geq \int_c^d \hat{0}(x) \ \mathrm{d} x = 0$$ by Theorem 6.12 (b) in Baby Rudin, where $\hat{0}$ denotes the zero function on $[c, d]$. That is, $$ \int_c^d f(x) \ \mathrm{d} x \geq 0 \tag{0}$$ for any points $c$ and $d$ such that $a \leq c \leq d \leq b$.

Suppose that there is a point $p \in [a, b]$ such that $f(p) > 0$. Then, as $f$ is continuous at $p$, so, for any real number $\varepsilon$ such that $$ 0 < \varepsilon < \frac{ f(p) }{2}, $$ we can find a real number $\delta > 0$ such that $$ \lvert f(x) - f(p) \rvert < \varepsilon < \frac{ f(p) }{2} $$ for all $x \in [a, b]$ for which $ \lvert x-p \rvert < \delta$.

Therefore, we can conclude that $$ \frac{ f(p) }{2} < f(x) < \frac{3 f(p) }{2} \tag{1} $$ for all $x \in I$, where $$I \colon= [a, b] \cap \left[ p- \frac{\delta}{2}, p+ \frac{\delta}{2} \right]. $$ Let's put $I \colon= [u, v]$. Then we see that $$ \begin{align} \int_a^b f(x) \ \mathrm{d} x &= \int_a^u f(x) \ \mathrm{d} x + \int_u^v f(x) \ \mathrm{d} x + \int_v^b f(x) \ \mathrm{d} x \\ & \qquad \qquad \mbox{ [ by an extension of Theorem 6.12 (c) in Baby Rudin ] } \\ &\geq \int_u^v f(x) \ \mathrm{d} x \qquad \mbox{ [ by (0) above ] } \\ &\geq \int_u^v \frac{f(p)}{2} \ \mathrm{d} x \qquad \mbox{ [ using (1) and Theorem 6.12 (b) in Baby Rudin ] } \\ &= \frac{f(p)}{2} (v-u) \\ &> 0, \qquad \mbox{ [ by (2) and our choice of $u$ and $v$ above ] } \end{align} $$ which contradicts our hypothesis that $\int_a^b f(x) \ \mathrm{d} x = 0$.

Is this proof sound enough in terms of its logic and rigor? If so, then is it also lucid enough in its presentation?

  • $\begingroup$ Your proof seems great. I would just suggest to avoid the introduction of $\epsilon$ which is not necessary. $\endgroup$ – mathcounterexamples.net Jul 26 '17 at 16:43
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    $\begingroup$ This should be the proof of the contrapositive, not a proof by contradiction. $\endgroup$ – egreg Jul 26 '17 at 17:40
  • $\begingroup$ @egreg The author of this proof is proceeding by contradiction; namely assuming that there is a point $p$ where $f(p)>0$, and then deducing a contradiction of what we assumed about the integral of $f$. $\endgroup$ – Alex Ortiz Jul 26 '17 at 18:52
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    $\begingroup$ @AOrtiz That's exactly what I objected to: the proof is unnecessarily by contradiction. If $f$ takes on a positive value, the integral is positive. $\endgroup$ – egreg Jul 26 '17 at 19:26

Your proof is fine. A little wordy and perhaps using more notation than you need, but it's correct.

I would have phrased it thusly:

Proof. Suppose $c\in [a,b]$ is such that $f(c)>0$. By continuity of $f$, there is a closed interval $[r,s]$ of $c$ where $f>f(c)/2>0$. By monotonicity of the integral, $$ \int_a^bf \ge \int_r^sf\ge \int_r^sf(c)/2>0, $$ which is a contradiction by what we assumed about $\int_a^b f$. $\qquad\square$


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