# Thread: Proof involving integrals equal to 0.

1. ## Proof involving integrals equal to 0.

Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

I believe I understand why it is true, I'm just not sure how to write the formal proof.

2. Originally Posted by amm345 Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

I believe I understand why it is true, I'm just not sure how to write the formal proof.

What have you tried? Here is a weird approach.

Suppose that $\displaystyle f\ne 0$ then there exists some $\displaystyle x\in[a,b]$ such $\displaystyle f(x)\in(0,\infty)$ and since this set is open there exists some open ball $\displaystyle B_{\varepsilon}(f(x))\subseteq(0,\infty)$ and by continuity there exists some $\displaystyle B_{\delta}(x)$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq(0,\infty)$. In other words, there exists some open ball around the point where $\displaystyle f(x)\ne0$ which is entirely positive. Take the concentric closed ball $\displaystyle B_{\frac{\delta}{2}}$. This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since $\displaystyle f:B_{\frac{\delta}{2}}[x]\mapsto\mathbb{R}$ is continuous we know that $\displaystyle f$ assumes a minimum on $\displaystyle B_{\frac{\delta}{2}}[x]$. In particular, $\displaystyle \inf_{x\in B_{\frac{\delta}{2}}[x]}f(x)=\varepsilon>0$. Thus, $\displaystyle \int_a^b\text{ }f=\int_a^{x-\frac{\delta}{2}}\text{ }f+\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}\text{ }f+\int_{x+\frac{\delta}{2}}^b \text{ }f\geqslant 0+\frac{\varepsilon \delta}{2}+0>0$. This of course is a contradiction.

3. That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.

4. Originally Posted by amm345 That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.
Which theorem is that?

5. Oh sorry, the Heine-Borel theorem.

6. Originally Posted by amm345 Oh sorry, the Heine-Borel theorem.
Are you serious? What's the point of having tools if you can't use them! What if I proved it first? 7. I have no idea.. I couldn't think of anything else it is kinda ridiculous!

8. Originally Posted by amm345 I have no idea.. I couldn't think of anything else it is kinda ridiculous!
Ok. I can think of one or two other ways....what are you currently doing in class?

9. We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.

10. Originally Posted by amm345 We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.
Hmm...Ok. I see an angle. Make a modified argument of mine and use the FTC.

11. Hello,

Why can't we just say that (without HB theorem) :
if there exists $\displaystyle m\in[a,b]$ such that $\displaystyle f(m)>0$, then by definition of the continuity, $\displaystyle \forall \epsilon>0, \exists \delta>0,\forall x\in[a,b], (|x-m|<\delta)\Rightarrow (|f(x)-f(m)|<\epsilon)$

In particular, if we let $\displaystyle \epsilon=\frac{f(m)}{2}$, $\displaystyle |f(x)-f(m)|<\epsilon \Leftrightarrow -\tfrac{f(m)}{2}<f(x)-f(m)<\tfrac{f(m)}{2} \Leftrightarrow \tfrac{f(m)}{2}<f(x)<f(m)$

So there exists $\displaystyle \delta>0$ such that $\displaystyle \forall x\in]m-\delta,m+\delta[$, $\displaystyle f(x)>\tfrac{f(m)}{2}>0$

Hence $\displaystyle \int_a^b f(x) ~dx=\int_a^{m-\delta} \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m+\delta}^b \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m-\delta}^{m+\delta} \underbrace{f(x)}_{> \frac{f(m)}{2}} ~dx$

So $\displaystyle \int_a^b f(x) ~dx>0+0+2\delta \cdot\tfrac{f(m)}{2}=\delta f(m)>0$

which is a contradiction.

So m can't exist, and hence $\displaystyle f(x)=0 ~,~ \forall x\in[a,b]$

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