Showing that $Unif(0,\theta)$ is complete.

Question

There is a part that I do not understand in the following proof so I would like to have some assistance.

Let $u(X)$ be a statistic where $X \sim Unif(0,\theta)$.

I know that I am to show that

$$E[u(X)]= 0 \quad \text{iff} \quad u(X)=0$$ .

So, we start the set up as

$$\int_0^{\theta} \frac{u(x)}{\theta} dx =0$$

Here the work that I am seeing is that you take the derivative with respect to $\theta$ on both sides leading to the following.

$$\frac{u(\theta)}{\theta} - \frac{1}{\theta^2}\int_0^{\theta} u(x) dx = 0$$

The part that I do not understand is that supposedly the expression

$$\int_0^{\theta} u(x) dx=0$$

I cannot think of any reason why this is true, so I would really appreciate your help.

Answer 1

$Eu(X)=\frac 1 {\theta} \int_0^{\theta} u(x)dx$. So $Eu(X)=0$ implies $\int_0^{\theta} u(x)dx=0$.