Conside $(X, \Gamma, \mu)$ to be a measure space. Define $\nu(A) = \mu(A)^2$ for any $A \in \Gamma$, where $\Gamma$ is a topology. Is this $\nu$ a measure on $(X, \Gamma)$?
This is what I've done: For $\nu$ to be a measure, we need $\nu(E)=\nu(E \cap A) + \nu(E \cap A^c)$. However, $\nu(E) = \mu(E)^2 = (\mu(E \cap A) + \mu(E \cap A^c))^2 = \mu(E \cap A)^2 + 2\mu(E \cap A)\mu(E \cap A^c) + \mu(E \cap A^c)^2$ which is not $\nu(E \cap A) + \nu(E \cap A^c)$ unless $\mu(E)=0$ $\forall E \in X$.
Does this seem correct or am I completely on the wrong track? Thanks!
Suppose $\nu$ is a (finite, non-zero) measure and consider some $A\in \Gamma$. Note that $$\begin{align} [\mu(A)+\mu(A^c)]^2 &=\mu(A\cup A^c)^2\\ &= \nu(A\cup A^c) \\ &= \nu(A)+\nu(A^c)\\ &= \mu(A)^2+\mu(A^c)^2 \end{align}$$ Hence $\mu(A)\mu(A^c)=0$ and we have $\mu(A)=0$ or $\mu(A^c)=0$.
If $\mu(A)>0$, we have $\mu(A^c)=0$ thus $\mu(A)=\mu(X)$. Hence $\mu$ takes values in the set $\{0,\mu(X)\}$.
If $A$ and $B$ are such that $\mu(A)>0$ and $\mu(B)>0$, then $\mu(A\cap B)=\mu(A)=\mu(B)=\mu(X)$, so $\mu$ must be quite pathological...