Theorem Reminder: Integral of $\phi$ Over [0,1]

[Bachelier]

Could someone remind me what theorem is this:

$let \ \phi: [0,1] \rightarrow \mathbb{R} \ , \ \phi \geqslant 0 \ , \ \phi \in \mathbb{R} \\[20pt] if \ \ \int_0^1 \mathrm{\phi(t)} \ \mathrm{d}t = 0 \ \Rightarrow \ \phi \equiv 0$

Thanks

 

[jbunniii]

I assume you mean $\phi \geq 0$, not $\phi > 0$ (else the conclusion $\phi = 0$ would be impossible). It's false without further assumptions. Let $\phi$ be the indicator function for any finite subset of $[0,1]$ for a counterexample. Do you mean to assume $\phi$ is continuous?

 

[pwsnafu]

I assume you mean $\phi \geq 0$, not $\phi > 0$ (else the conclusion $\phi = 0$ would be impossible). It's false without further assumptions. Let $\phi$ be the indicator function for any finite subset of $[0,1]$ for a counterexample. Do you mean to assume $\phi$ is continuous?

[strike]He's written $\phi \equiv 0$ so he probably means a.e.[/strike]

Maybe not. But yeah equality is only up to almost everywhere.

 

[jbunniii]

[strike]He's written $\phi \equiv 0$ so he probably means a.e.[/strike]

Maybe not. But yeah equality is only up to almost everywhere.

Good point. OP, can you give us more context? Is it a Riemann integral or a Lebesgue integral?

 

[Bachelier]

I assume you mean $\phi \geq 0$, not $\phi > 0$ (else the conclusion $\phi = 0$ would be impossible). It's false without further assumptions. Let $\phi$ be the indicator function for any finite subset of $[0,1]$ for a counterexample. Do you mean to assume $\phi$ is continuous?

fixed. $\phi \geqslant 0, \ \phi$ is continuous.

 

[Bachelier]

Good point. OP, can you give us more context? Is it a Riemann integral or a Lebesgue integral?

Riemann integral

 

[jbunniii]

fixed. $\phi \geqslant 0, \ \phi$ is continuous.

OK, I don't know if the theorem has a name, but it's easy to prove. If $\phi$ is not identically zero, take a point $x$ where $\phi(x) > 0$. Then there is a neighborhood of positive radius around $x$ where $\phi(y) > \phi(x)/2$ for all $y$ in the neighborhood. So the integral is at least $\phi(x)/2$ times the width of the neighborhood.

 

[Bachelier]

Then there is a neighborhood of positive radius around $x$ where $\phi(y) > \phi(x)/2$ for all $y$ in the neighborhood.

Do we get this because of the continuity of $\phi$?

So the integral is at least $\phi(x)/2$ times the width of the neighborhood.

did you mean then that the integral is at least $\epsilon \cdot \frac{\phi(x)}{2}$ Hence $\equiv$ to 0?

 

[jbunniii]

Do we get this because of the continuity of $\phi$?

Yes.

did you mean then that the integral is at least $\epsilon \cdot \frac{\phi(x)}{2}$ Hence $\equiv$ to 0?

It's at least $\epsilon \cdot \frac{\phi(x)}{2}$ (assuming $\epsilon$ is the width of the neighborhood described earlier), so the integral is NOT zero.

To summarize, we proved that if $\phi$ is continuous and $\phi(x) > 0$ for some $x$, then $\int \phi > 0$. Equivalently, if $\int \phi = 0$, then there cannot be any $x$ for which $\phi(x) > 0$, i.e., $\phi$ must be identically zero.