If $\int_a^b u(x) dx=0$ then $Pr[u(X)=0]=1$?

Question

We are given that $X \sim Unif(a,b)$ and $u(X)$ is a statistic.

I do not understand why

$\int_a^b u(x) dx=0$ then $Pr[u(X)=0]=1$

What I can conclude is that

$$\int_0^a u(x)dx = \int_0^b u(x)dx$$ but I am not sure how that connects to the probability expression on the right side.

What bothers me most is the fact that the probability is not an interval so I do not know how the probability is not actually $0$.

$Edit$

Thank you for your help.

It seems like the information I have provided was not sufficient so I will show everything that I was given.

$X \sim Unif(a,b)$, $u(X)$ is an unbiased and sufficient statistic.

My goal is to prove that $X$ is a complete distribution, so I must show

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

This leads to the equation I showed above, but I am not sure how it connects to the probability statement.

I also considered showing that $X$ is in the regular exponential class but since the support depends on $a$ and $b$ I could not do that.

Answer 1

$0=\int_a^b u(x)dx=\mathbb E[u(X)]$. So unless if $\text{Var}(u(X))=0$, the statement is not correct.

Answer 2

The statement is true if $u$ has the same sign almost everywhere on $[a,b]$. Otherwise it's quite possible to have cancellation. For example, try $a=0$, $b=1$ with $u(x) = 2x-1$.