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

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

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

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

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

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