Subspace of a measure space. Let $(X,\mu)$ be a measure space and let $Y \subset X$ be $\mu$-measurable with $\mu(Y) > 0$. Define an outer measure $\nu$ on $Y$ by putting $\nu(S) = \mu(S)$ for $S \subseteq Y$.
I need to show that if $S \subseteq Y$ is $\nu$-measurable as a subset of $Y$, then it is $\mu$-measurable as a subset of $X$. Recall that S is $\mu$-measurable if for all $T \subset X$ $$\mu(T) = \mu(T \cap S) + \mu(T \cap (X \setminus S)).$$ In doing this, I get the following:
Let $T \subseteq X$. Then $$\mu(T \cap Y) = \nu(T \cap Y) = \nu(T \cap S) + \nu(T \cap (Y \setminus S)) = \mu(T \cap S) + \mu(T \cap (Y \setminus S))$$ and $\mu(T) = \mu(T \cap Y) + \mu(T \cap (X \setminus Y))$, so that $$\mu(T) = \mu(T \cap S) + \mu(T \cap (Y \setminus S)) + \mu(T \cap (X \setminus Y)).$$
This almost solves it, but I don't know that $Y \setminus S$ is $\mu$-measurable, so I can't simply write that $\mu(T \cap (Y \setminus S)) + \mu(T \cap (X \setminus Y)) = \mu(T \cap (X \setminus S))$ and I'm not sure that it's any easier to show this than it is to show that $S$ is itself $\mu$-measurable. Any suggestions. 

You know that $Y$ is $\mu$-measurable, and
$$T\cap (Y\setminus S) = (T\setminus S)\cap Y; \quad T\cap (X\setminus Y) = (T\setminus S)\setminus Y.$$
That immediately yields
$$\mu(T\setminus S) = \mu(T\cap (Y\setminus S)) + \mu(T\cap (X\setminus Y)).$$
