If $\nu$ and $\mu$ are finite measures on ${\cal E}$ such that $\nu(E) \le \mu(E)$ for each $E \in {\cal E}$ and $\nu(X)= \mu(X)$.

Question

I want to prove or disprove the following proposition. I believe it is false but I know very few examples of measures and cannot construct two examples with this property.

If $\nu$ and $\mu$ are finite measures on a $\sigma$-algebra ${\cal E}$ such that $\nu(E) \le \mu(E)$ for each $E \in {\cal E}$ and $\nu(X)= \mu(X)$, then $\nu = \mu$ as functions.

Answer 1

$\nu (E^{c}) \leq \mu (E^{c})$ so $\nu (X) -\nu (E) \leq \mu (X) -\mu (E)$ which gives $\mu(E) \leq \nu (E)$. Hence $\mu =\nu$.