I}$ in $\Omega$ and a countable set $\{\mu_i\}_{i \in I}$ of positive real numbers such that $$\mu = \sum_{i \in I} \mu_i \delta_{x_i}.$$

We divide the proof into two steps.\
[**Step 1.**]{} Assume that $\mu = \nu$, choose any measurable set $A$, and assume that (\[hypothesis\]) holds; by Lemma \[aproksymacja\] $$\| \mathbf{1}_{A} \|_{L^{r}(\Omega,\mu)} \leq C \| \mathbf{1}_{A} \|_{L^{p}(\Omega,\mu)},$$ and then $$\mu(A)^{\frac{1}{r}} = \| \mathbf{1}_{A} \|_{L^r(\Omega,\mu)} \leq C \| \mathbf{1}_{A} \|_{L^p(\Omega,\mu)} = C \mu(A)^{\frac{1}{p}}.$$ Hence either $\mu(A) = 0$, or $\mu(A) \geq 1 / {C}^{\frac{pr}{r- p}}$. Then by Lemma \[lemma:delty\] there exists finite set $\{x_i \}_{i \in I}$ of points in $\Omega$ and a finite set $\{ \mu_i \}_{i \in I}$ of real numbers with $\mu_i \geq  1/C^{\frac{pr}{r- p}}$ such that $$\mu = \sum_{i \in I} \mu_i \delta_{x_i}.$$

[**Step 2.**]{} Now assume that $\mu$ and $\nu$ are arbitrary; the Lebesgue Decomposition theorem ensures that $$\label{decomposition}
\nu = \mu \llcorner \theta + \sigma$$ for some non-negative $\theta$ in $L^1(\Omega,\mu)$, where $\mu \llcorner \theta (A) := \int_{A} \theta d \mu$, and $\sigma$ is a positive measure singular with respect to $\mu$. For every positive integer $n$ consider the function $$\phi_n := \theta^{\frac{1}{r - p}} \mathbf{1}_{\theta \leq n} \psi   ,$$ where $ \psi $ 