Let $\mu,\nu$ be $\sigma$-finite on $(\Omega,\mathcal{A})$ and $\nu \le \mu$. Then there exists a $\mu$-almost-everywhere unique function $f=\frac{d\nu}{d\mu}$ with $0 \le f \le 1 \ \mu \ a.e.$
As both measures are $\sigma$ finite and $\nu \le \mu \Rightarrow \nu << \mu$ the Radon-Nikodym-Theorem tell us that there exists such a function $f$ which fulfils $f\ge 0$ $\mu \ a.e.$ and this $f$ is unique $a.e.$
The only thing that is left to show is that $f \le 1 \ a.e.$. For this I tried a proof by contradiction:
Let $A \in \mathcal A$ with $\mu (A)>0$ and $\forall x \in A:f(x)>1$
Then $\nu(A)=\int_A f d \mu>\int_A 1 d \mu=\mu(A)$ which is a contradiction because $\nu \le \mu$ and therefore $f \le 1 \ a.e$.
Is my way to show $f \le1$ correct?
Well, I presume $\nu\leq \mu$ means $\nu(A) \leq \mu(A)$ for each $A\in \mathcal A$. In that case, let's assume the contrary to the statement you need to prove, i.e. $\mu(B) > 0$ where $B = \{f > 1\}$. We get $$ \nu(B) - \mu(B) = \int_B(f - 1)\mathrm d\mu > 0 $$ which contradicts the fact that $\nu\leq \mu$. Now, the fact that a Lebesuge integral of a function positive over a set of positive measure is positive I hope you can show yourself. As a hint: split this set into subsets of $\left\{f\geq\frac1n\right\}$ where $n$ runs over positive integer numbers.