This is problem 6.2 from the 3rd edition of Principles of Mathematical Analysis.

Problem 6.2: Suppose $f\geq 0$, f is continuous on $[a, b]$, and $\int_a^b f(x) \, dx = 0$. Prove that $f(x)=0$ for all $x \in [a, b]$.

I'm looking for a critique of my proof. It's a pretty easy problem, but I am always wary of making too bold of assumptions, especially on these low level/fundamental proofs. I'll be using Rudin's notation and refer to theorems from the text (If I should include the text of each theorem, feel free to leave a comment... I'm lazy but could probably use the TeX practice :p)

Proof: Assume, for contradiction, that $f>0$. Then, for any partition $P$ we have the Lower Riemann Sum: $L(P, f)=\sum_{i=1}^n m_i \, \Delta x_i$. At least one $\Delta x_i$ must be positive, since $a < b$, and each $m_i$ must be positive since we have $f>0$ by assumption, so certainly $\sup f > 0$. That means $L(P, f)>0$. Thus, we have: $$0 = 0(b-a) < L(P,f)\leq \sup L(P, f) = \inf U(P,f) =L$$ where the last string of equalities holds because our function is continuous on a compact interval, so is integrable by theorems 6.8 and 6.6. So our integral has value $L>0$. This is in contradiction to our given assumption that $\int_a^b f(x) \, dx = 0$, so we must have that $f=0$ on $[a, b]$. $\Box$

So, I am wondering if my proof is correct (and is presented well). Also if someone could enlighten me as to what Rudin means when he say "Compare this with exercise 1," I'd be appreciative. Is it a hint or is there something else he expects you to notice? there are a lot of things I could compare :)...

Exercise 6.1: Suppose $\alpha$ increases on $[a, b]$, $a \leq x_0 \leq b$, $\alpha$ is continuous at $x_0$, $f(x_0)=1$, and $f(x)=0$ if $x\neq x_0$. Prove that $f$ is Riemann-Stieltjes Integrable and that $\int f \, d\alpha = 0$

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    $\begingroup$ Perhaps you should include the problem statement too? $\endgroup$
    – Aryabhata
    Jan 26, 2012 at 0:41
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    $\begingroup$ Good point! I edited the question to include the problem statement. $\endgroup$
    – Tyler
    Jan 26, 2012 at 1:02
  • $\begingroup$ I don't think you should include the statements all the theorems you use in the proof. However, you should definitely include the statement of "exercise 1" since you ask us to compare this to it. $\endgroup$
    – Srivatsan
    Jan 26, 2012 at 1:06

4 Answers 4


A proof: If $f\ge0$ everywhere and $f(x_0)>0$ and $f$ is continuous, we could do a little $\varepsilon$-$\delta$ argument like this: Let $\varepsilon=f(x_0)/2$. Let $\delta>0$ be small enough so that if $x$ is within distance $\delta$ of $x_0$, then $f(x)$ is within $\varepsilon$ of $f(x_0)$. So $f\ge f(x_0)/2$ on the interval whose endpoints are $x_0 \pm \delta$, and so $$ \int_a^b f(x) \; dx \ge (2\delta) (f(x_0)/2) = \delta f(x_0) > 0. $$

To allow for $x_0$ being near an endpoint, you could just integrate over half that interval.

Some comments on the posted proof: The assumption in a proof by contradiction should not be stated as "$f>0$". Rather it should be stated as saying there is at least one point $x_0$ such that $f(x_0)>0$. Whenever the conclusion says "All A are B", then the assumption in a proof by contradiction should be "At least one A is not B".

Your argument to the conclusion that $\sup f>0$ is too complicated. If you've assumed $f$ is not everywhere $0$ and you have $f\ge 0$ everywhere, then as soon as you've assumed there is one point $x_0$ where $f>0$, you've already got $\sup f\ge f(x_0)$.

Since you're working with Riemann integrals defined by Riemann sums, you might make the partition $\{a, x_0-\delta,x_0+\delta, b\}$ and then you have the lower Riemann sum $\ge (2\delta) (f(x_0)/2)$. If the lower Riemann sum for just one partition is positive, then the integral is positive.

Saying "Then for any partition..." seems at best a needless complication. Just one partition, as noted above, is enough if you do the right things with it.

  • $\begingroup$ Thanks. The answers were all helpful, but this one compiled all the remarks, as well as clearing up my over-complications and addressing the question of partitions as well. $\endgroup$
    – Tyler
    Jan 26, 2012 at 4:36

Your proof is incorrect.

You need to show that at every point $x \in [a,b]$, $f(x) = 0$.

To argue by contradiction, you need to assume, there is at least one $c \in [a,b]$ such that $f(c) \gt 0$.

You are assuming that $f(x) \gt 0 \ \forall x \in [a,b]$ and proving that is false.

In effect, you have only shown that there is at least one point $c \in [a,b]$ for which $f(c) = 0$.

  • $\begingroup$ The inequality is not helping his argument. $\endgroup$ Jan 26, 2012 at 1:24
  • $\begingroup$ @ncmathsadist: Sorry, I don't understand what you mean by that. $\endgroup$
    – Aryabhata
    Jan 26, 2012 at 1:30
  • $\begingroup$ You are using the assumption that $f > 0$ throughout the interval to establish the displayed inequality and that is a flawed premise. So the inequality is not going to help you. $\endgroup$ Jan 26, 2012 at 1:35
  • $\begingroup$ @ncmathsadist: I am not assuming anything. All I am doing is pointing out the flaw in his proof, as he has requested. $\endgroup$
    – Aryabhata
    Jan 26, 2012 at 1:37
  • $\begingroup$ Sorry for the confusion; my comment was aimed at the author, not at you, Aryabhata. $\endgroup$ Jan 26, 2012 at 1:46

You do not need an argument by contradiction. You can use the contraposititive instead. Suppose that at some $x_0\in [a,b]$, we have $f(x_0) > 0$. Then, by continuity, there is an interval containing $x_0$ on which $f(x) \ge f(x_0)/2$. Define a function $g$ which is zero off the interval and $f(x_0)/2$ on the interval. Let $\delta$ be the length of the interval. We have $f \ge g$ so $$\int_a^b f(x)\,dx \ge \delta f(x_0)/2 > 0.$$

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    $\begingroup$ How much difference is there between proving the contrapositive of "If A then B", and proving "If A then B" by contradiction? I'd have said they're the same thing. $\endgroup$ Jan 26, 2012 at 2:15
  • $\begingroup$ @MichaelHardy: Thanks for your comment. I was confused at how this was a proof by contrapositive, but now it makes sense (negating the assumptions creates a string of OR's... then we only need to prove one of the statements... choose the integral one, and assume the other two original statements since their truth values are immaterial to the truth value of the entire statement). Uff da. Thinking of it as a proof by contradiction seems much simpler, despite them being equivalent. $\endgroup$
    – Tyler
    Jan 26, 2012 at 2:30

You're plenty of advices for how to prove that by contradiction, but some others mistakes must be pointed out. For example, $\Delta x_i=x_i-x_{i-1}\gt 0$ for $i\in\{1,\ldots,n\}$ and $m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$ so, you must argue: since $f\gt 0$ certainly $\inf f\gt 0$...

However, it's possible to prove this directly. You can argue as follows.

Since $f\geq 0$ everywhere in $[a,b]$, the function $F:[a,b]\to\mathbb{R}$, given by $$F(x):=\int_a^x f(t)\, dt,$$ is increasing. Then you have $$0=F(a)\leq F(x)\leq F(b)=0.$$ Therefore $F$ is the constant function $0$. Now, since $f$ is continuous in $[a,b]$ by the fundamental theorem of calculus, we have $$f(x)=F'(x)=0.$$

  • $\begingroup$ The partition is defined with $x_i \leq x_{i+1}$, so should we say "WLOG, $\Delta{x_i} > 0$"? Mixing up the sups and infs in the definition of the Riemann Sums was a stupid mistake on my part! Thank you. I was actually trying to prove that at another place in my proof sketch before I wrote this up! I guess I could also argue similarly to the other posts (and so avoid using the FTC) by saying "that if $f(x_0) > 0$ then by continuity $f(x) > f(x_o)/2$ on an interval... so the $\inf{f} > 0$ on that interval. $\endgroup$
    – Tyler
    Jan 26, 2012 at 16:40
  • $\begingroup$ I was trying to struggle through the definitions and get my hands dirty to solve the problem... but I am definitely happy to see such an elegant solution! $\endgroup$
    – Tyler
    Jan 26, 2012 at 16:42
  • $\begingroup$ @TylerBailey, thanks I'm glad to help you. Yes is correct to say: "WLOG...". I did say: "... $\Delta x_i\gt 0$ for all $i$..." because I was thinking that your partitions have the form $$P=\{a=x_0\lt x_1\lt\ldots\lt x_n=b\}.$$ Another way to see that if $f\gt 0$ then, $\inf f\gt 0$ over any $[x_{i-1},x_i]$, is by contradiction. Suppose that there is a $j$ such that $\inf_{x\in [x_{j-1},x_j]} f(x)=0$. Since $f$ is continuous in $[x_{j-1},x_j]$, which is compact, attains a minimum. Suppose that this minimum is attained in $a\in [x_{j-1},x_j]$. Then $f(a)=0$, and this contradict that $f\gt 0$. $\endgroup$
    – leo
    Jan 30, 2012 at 19:54

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