How to conclude that if the integration of function $f$ over arbitrary domain $= 0$, then $f = 0$ too? I am engineer and not so deep in math and i need your help please.
Assume that $f (x,y,z)$ is any continuous function and  that
$\int_{\Omega} f(x,y,z) \ {\rm d}v = 0$
If $\Omega$ is to be any arbitrary bounded domain, then
$f = 0$
How to prove this statement?
Maybe we can use the fact that if f is strictly positive then this is true and then we can divide our domain into regions where function is strictly positive  and regions where f is strictly negative? any idea?
 A: Proceeding by contradiction:
Suppose $f(a,b,c)=A>0$. Let $0<\delta<A$. Then by the definition of continuity, there is an open ball $B$ of radius $\varepsilon>0$ around $(a,b,c)$ on which $\lvert f(x,y,z)-A \rvert<\delta$, or in other words,
$$ f(x,y,z)>A-\delta>0 $$
for every $(x,y,z) \in B$. Now integrate the above inequality over $B$:
$$ \int_B  f(x,y,z) \, dV \geqslant (A-\delta)\operatorname{Vol(B)} >0, \tag{1} $$
where $\operatorname{Vol(B)}$ is the (positive) volume of $B$. The first inequality can become non-strict, but the second cannot, since the integral of a constant function over a set of positive area is positive (think about rectangles). (1) is then what you want: we have found a set where the integral of $f$ is not zero. Hence $f$ cannot have a point where it is positive (or negative, if we do the same idea for $-f$), and so $f \equiv 0$.
A: Hint. Assume not. Assume there is a point $\vec{x}_0$ with $f(\vec{x}_0)\neq 0$ and WLOG assume $f(\vec{x}_0)> 0$ . Use continuity to find an region $\Omega \ni \vec{x}_0$ where $f(\Omega)>0$ (hint: let $\epsilon = f(\vec{x}_0)/2$). Then use a comparison theorem for integrals to arrive at a contradiction.
A: You have the right idea.  Consider $\Omega$ of the form 
$$\{(x,y,y)\in\Bbb R^3: f(x,y,x)>k, x^2+y^2+z^2 <M\},
$$
for various $k>0$ and (large) $M>0$. For such an $\Omega$,
$$
0=\int_\Omega f\,d\nu\ge k\cdot\nu(\Omega).
$$
This forces the  open set $\Omega$ to have zero volume, hence to be empty.
Collecting this information as $k>0$ and $M>0$ vary, we see that $f(x,y,z)$ cannot ever be $>0$. A similar argument shows that $f$ can't take negative values.
