# Prob. 2, Chap. 6, in Baby Rudin: If $f\geq 0$ and continuous on $$with \int_a^bf(x)\ \mathrm{d}x=0, then f=0 Here is Prob. 2, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: Suppose f \geq 0, f is continuous on [a, b], and \int_a^b f(x) \ \mathrm{d} x = 0. Prove that f(x) = 0 for all x \in [a, b]. My Attempt: Let c and d be any points in [a, b] such that c \leq d. As f(x) \geq 0 for all x \in [a, b] and as f is continuous on [a, b], so f(x) \geq 0 for all x \in [c, d] and f is continuous on [c, d]; therefore f \in \mathscr{R} on [c, d] and$$ \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}
$$which contradicts our hypothesis that \int_a^b f(x) \ \mathrm{d} x = 0. Is this proof sound enough in terms of its logic and rigor? If so, then is it also lucid enough in its presentation? #### Solutions Collecting From Web of "Prob. 2, Chap. 6, in Baby Rudin: If f\geq 0 and continuous on$$ with $\int_a^bf(x)\ \mathrm{d}x=0$, then $f=0$"

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 $\int_a^b f$. $\qquad\square$