Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$

Question

Let $a<b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$

Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D> 0$

Since the function is continuous at $x_0$, we have $$\forall \epsilon>0, \exists \delta>0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2>0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f >0$

Now since $\int_{[a,b]} f = 0$. we have

$$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f<-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$

Is my proof correct?

Answer 1

Your proof looks completely right to me. In fact, it's the one I was planning to write if the question hadn't included a proof.

The only thing I'd change is to say that the integral, over the reduced interval, is at least $\frac{D}{2} 2\delta = \delta D > 0$. It's always nice to see a concrete estimate like that.

I might also simplify a little and start out by observing that $\int_{[0, a]} f = 0$ for any $0 < a < 1$ because $f$ is nonnegative, so the integrals from $0$ to $a$ and $a$ to $1$ must be nonnegative, but they sum to zero, and hence are both zero. Further such reasoining shows that $\int_I f = 0$ for any subtinterval of $[0, 1]$.

Once you've done that, you can stop your proof just before "This implies" by just saying "and now we've exhibited an interval over which the integral is positive, a contradiction."

Answer 2

Yes, I think there's no problem with your solution. We can say in short: since $f\ge 0$, its graph lies above $Ox$. If $f(x_0)>0$ for some $x_0$, then there is a neighborhood of $x_0$ such that for all $x_1<x_2$ in that neighborhood, we have $f(x_1)>0,f(x_2)>0$. Then the area under the curve on the segment $[x_1,x_2]$ is greater than $0$, which is impossible.

Answer 3

This isn't exactly what you're asking for, but this is how I would approach this problem:

If $f$ is Riemann integrable, then it is Lebesgue integrable and the two integrals are equal. So if $\int_{[a,b]}f\mathrm dm=0$ where $m$ is Lebesgue measure, then $f=0$ almost everywhere (since $f$ is nonnegative). Since $f$ is continuous, it follows that $f$ is identically zero.