$\left|\int g(t)dt\right|=0$ then can we say that $g=0$? Let $g$ be a continuous function on $[0,1]$ such that
$$\left|\int g(t)dt\right|=0$$
then can we say that $g=0$?
since $g$ is continuous, so it is measurable. 
from this can we have this result?
please help me.
 A: From your comment to your question, the statement should be:
Let $g:[0,1]\to\mathbb{R}$ be continuous and such that for all $x,y\in[0,1]$ we have:
$$\left|\int_x^yg(t)dt\right|=0$$
Is it necessarily true that $g(x)=0$.
The answer to this question is yes.
Indeed assume there was some $x_0\in[0,1]$ with $g(x_0)\neq 0$, then (by continuity) there would be a little closed neighborhood $x_0\in[x_0-\epsilon,x_0+\epsilon]$ where $g$ is never zero (and thus is always positive, or negative, depending on the sign of $g(x_0))$. Then:
$$\left|\int_{x_0-\epsilon}^{x_0+\epsilon}g(t)dt\right|>0$$
A: Your condition, $$\left|\int g(t)dt\right| = 0,$$
is equivalent to $$\int g(t)dt = 0.$$
This condition does not require a function to equal zero. Since you demand that only the integral of the function is zero, there are many functions that cover this. In fact, take any function $g(t)$. Now take the function $G(t) = g(t) - C$ where $C=\int_0^1g(\tau)d\tau$ (a constant). Look at the integral of $G$:
$$\int_0^1 G(t)dt = \int_0^1 g(t)dt - \int_0^1Cdt = \int_0^1g(t)dt - C = \int_0^1g(t)dt - \int_0^1g(\tau)d\tau=0.$$
Therefore, there are many many many functions that integrate to $0$ on $[0,1]$.
A much stronger conclusion can be made if $\int|g(t)|dt = 0$. In that case, $g(t)=0$ for almost all $t\in[0,1]$ and, if $g$ is continuous, $g(t)=0$.
