Show this set has measure zero

Question

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.

Answer 1

These, the OP already proved. I just add this for to get the main ideas to solve the issue in which the OP is interested.

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 $\mu(\{f=1\})=0$.

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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.