If a measure $\mu$ is less than a measure $\nu$ on a generating $\pi$-system, can we conclude that $\mu \leq \nu$? Let $\mu$, $\nu$ be finite measures on the non-degenerate compact interval $[a, b] \subseteq \mathbb{R}$ provided with the Borel $\sigma$-algebra. It is well-known that if $\mu(B) = \nu(B)$ for every Borel set $B$ belonging to a $\pi$-system that generates $\mathcal{B}([a, b])$ and that includes $[a, b]$, then $\mu = \nu$. Is the same true if the equality is replaced by an inequality? In other words, suppose $\mu(B) \leq \nu(B)$ for every Borel set $B$ belonging to a $\pi$-system that generates $\mathcal{B}([a, b])$ and that includes $[a, b]$. Is it the case that $\mu \leq \nu$?
 A: Just to elaborate a bit. First of all, the above example of Conrado works perfectly, but perhaps the following one is a bit easier to memorize, and it may clearly deliver the main idea that Conrado formulated in the last message of his answer.
Let $\Omega = \{a,b\}$ and $\mathcal C = (\{a\}, \Omega)$, then $\mathcal C$ is a $\pi$-system that generates $2^\Omega$. Define $\mu = (1,0)$ and $\nu = (0,1)$. Obviously, $\mu \geq \nu$ on $\mathcal C$, but it's not true that $\mu\geq \nu$ on $2^\Omega$.
Notice though that a similar result holds true. The difference is that the rectangles do not only form a $\pi$-system, they also have an important property that any element of the algebra that they generate can be written as a union of rectangles - hence $\mu\geq \nu$ on the whole algebra which is in turn enough for the inequality to hold over the $\sigma$-algebra. Clearly, in counterexample above (as in Conrado's) the algebra can't be written as a union of elements of the $\pi$-system, hence the inequality does not have to hold on the algebra. So
$$
  \mu \geq_{\mathcal C} \nu \quad \overset{?}{\implies} \quad  \mu \geq_{\mathcal A}\nu \quad \iff\quad \mu \geq_{\mathcal F} \nu \quad \implies \quad \mu \geq_{\mathcal C} \nu
$$
where $\mathcal C$ is a $\pi$-system that generates an algebra $\mathcal A$, which generates a $\sigma$-algebra $\mathcal F$. Of course, the middle equivalence is trivial in finite space since $\mathcal A = \mathcal F$, and the challenge of my question was to show it holds for the general case.
A: No, consider the following $\pi$ system:
$$\mathcal{C}=\{[a,t), [a,b]\; \vert \;t \in [a,b]\}$$
define
$\mu([a,t) = t-a + 100$ 
$\nu ([a,t)) = t-a + 50$ 
$\mu([a,b]) = 100 + (b - a) $
$\nu ([a,b]) = 100 +(b-a)$.
Note that  $\mu(C) \geq \nu(C) $ for every $C \in \mathcal{C}$, but $\mu (\{b\}) = \mu ([a,b]) - \lim_n \mu([a,b-1/n) = 0$ and $\nu (\{b\}) = 50$.
This is possible because we can't assure that $B \backslash A \in \mathcal{C}$ for a $\pi$- system
