# Show this set has measure zero

I have a measurable space $$(X, \Sigma)$$ with $$\mu, \nu: \Sigma \rightarrow [0, \infty)$$ finite measures. Let $$\lambda = \mu + \nu$$. Let $$f: X \rightarrow \mathbb{R}$$ be a $$\Sigma$$-measurable function such that $$\nu(E) = \int_E f\ d\lambda$$ for all $$E \in \Sigma$$.

In this problem, I was asked to first show that $$0 \leq f \leq 1$$ a.e. and $$\mu(f^{-1}(\{1\})) = 0$$, both of which I have successfully done. I'm struggling with the proving last part, which says that if $$A\subseteq \{x: 0 \leq f(x) < 1\}$$ and $$\mu(A) = 0$$, then $$\nu(A) = 0$$.

I know that if $$\mu(A) = 0$$, then $$\lambda(A) = \nu(A)$$. But we also have $$\nu(A) = \int_A f\ d\lambda$$. So $$\lambda(A) = \int_A f\ d\lambda$$. From here, I don't know where to go. I feel that $$\mu(f^{-1}(\{1\})) = 0$$ will come into play at some point, but I can't seem to incorporate it. I would appreciate any hints.

Since $$0\leq \nu\leq \lambda$$ and $$\nu=f\cdot\lambda$$, then $$0\leq f\leq 1$$ $$\lambda$$-a.s.
Since $$\lambda(\{f=1\}=\mu(\{f=1\})+\int_{\{f=1\}}f\,d\lambda =\mu(\{f=1\})+\lambda(\{f=1\})$$ it follows that $$\mu(\{f=1\})=0$$.
Now, for any set $$B$$ with $$\mu(B)=0$$, $$\lambda(B)=\nu(B)=\int_Bf\,d\lambda$$ hence $$\int_B(\mathbb{1}-f)\,d\lambda=0$$ Since $$0\leq f\leq1$$, it follows either (a) $$\lambda(B)=0$$ or (b) $$\lambda(B)>0$$ and $$(\mathbb{1}-f)\mathbb{1}_B=0$$ $$\lambda$$-a.s.
In particular, if $$A\subset\{0\leq f<1\}$$ and $$\mu(A)=0$$, then (b) can not hold so (a) holds.