Riemann integral of continuous non-negative function

Question

Suppose $f\geqslant 0$, $f$ is continuous on $[a,b]$, and $\int \limits_{a}^{b}f(x)dx=0.$ Prove that $f\equiv 0$ on $[a,b]$.

Proof: We'll define function $F(x)=\int \limits_{a}^{x}f(t)dt$. We know that $F(x)$ is continuous since $f$ is bounded and also $F(x)$ is differentiable since $f$ is continuous and $F'(x)=f(x)$ for any point $x\in [a,b]$.

Also $F(x)=0$ for any $x\in [a,b]$. Why? Because $$0=\int \limits_{a}^{b}f(t)dt=\int \limits_{a}^{x}f(t)dt+\int \limits_{x}^{b}f(t)dt$$ but both terms are non-negative. Hence $F(x)=0$ for any $x\in [a,b]$. Thus $F'(x)=f(x)=0$. Q.E.D.

What you think about that proof?

Something like: therefore $F(x)=-\int_x^b f(t)dt\leq 0$. Since $F(x)\geq 0$, we get $F(x)=0$ would have been better. — Surb, Nov 06 '15 at 10:42
One can also say that as $F^{\prime}(x)=f(x)\geq 0$, $F$ is increasing and $0=F(a)\leq F(x)\leq F(b)=0$, but your idea seems nicer. — Kelenner, Nov 06 '15 at 10:44
You might find it interesting to try to give a proof from the definition of the Riemann integral, without using the fundamental theorem of calculus. — Nate Eldredge, Nov 06 '15 at 14:43
@NateEldredge, What is fundamental theorem of calsulus? You means Newton-Leibniz theorem? I think that I don't use it in my proof. — RFZ, Nov 06 '15 at 15:11

score 5 · Accepted Answer · answered Nov 06 '15 at 10:45

5

An other way

Since $f\geq 0$, the function $F:x\mapsto \int_a^x f(t)dt$ is increasing on $[a,b]$. Therefore, if $x\in [a,b]$, $$0\leq F(x)\leq F(b)=0,$$ what prove the claim.

answered Nov 06 '15 at 10:45

Surb

55,662

Also nice way of proof! +1 – RFZ Nov 06 '15 at 10:54

score 1 · Answer 2 · answered Nov 06 '15 at 10:36

1

Your proof seems to be fine.

It is also possible, to give a proof by contradiction. Suppose $f(x) \ne 0$ for some $x \in [a,b]$....

answered Nov 06 '15 at 10:36

gerw

31,359

Yes you are right! I also proved it by contradiction if you want i can post it. But above proof seems more elegant to me. – RFZ Nov 06 '15 at 10:37

score 1 · Answer 3 · answered Nov 06 '15 at 12:35

The question is already answered, but I will give an alternative answer in terms of the Lebesgue theory. Since $f$ is continuous on $[a,b]$, it is Lebesgue integrable. As $f\geqslant0$ and $\int f\ \mathsf d\mu = 0$, $f=0$ a.e.

To see this, let $E_n = \left\{x: f(x)>\frac1n\right\}$. Then if $\mu(E_n)>0$ for any $n$, we have $$\int f\ \mathsf d\mu \geqslant \frac{\mu(E_n)}n > 0, $$ a contradiction.

Riemann integral of continuous non-negative function

3 Answers3