Find a sequence $(\phi_n)_n \subset C^{\infty}_c(\mathbb{R}^N)$ which converges in both $L^p(\nu)$ and $L^q(\mu)$ to $1_E$ $\newcommand{\R}{\mathbb{R}}$
$\newcommand{\Cr}{C^{\infty}_c(\R^N)}$
Suppose we have two non-zero Borel measures on $\R^N$, labeled $\nu$ and $\mu$, and we have $1 \leq p, q < \infty$. Let $E \subset \R^N$ be measurable.
Is it possible to find a sequence $(\phi_n)_n \subset \Cr$ such that which converges to the characteristic function $1_E$ in both $L^p(\nu)$ and $L^q(\mu)$?
Context
I am reading a paper by El Hamidi and Rakotoson, and I am confused about one of the lines of reasoning that they present. They construct two measures $\nu$ and $\mu$, and then prove the following result (lemma 5, p.745):

For every $\phi \in \Cr$, we have:
  \begin{align*}
\left( \int |\phi|^{p} d\nu \right)^{1/p}
\leq C \left( \int |\phi|^{q} d\mu \right)^{1/q}
\end{align*}

(Here, $C$ is a constant that depends on $\mu,\nu,p,q$, but not $\phi$, and the coefficients are renamed $p$ and $q$ for clarity).
Then, later on (proof of corollary 1 of lemma 5, p.749), they claim the following:

The [above inequality] implies that for all Borelian sets $E \subset \R^N$, one has:
  \begin{align*}
\nu(E)^p \leq C \mu(E)^q
\end{align*}

This result seems intuitive to me, but I can't figure out how to prove it. Since $\nu(E) = \int 1_E d\nu$ and similarly for $\mu$, it seems like the straightforward way to prove this would be to approximate $1_E$ by smooth functions, then take the limit. However, this would require an approximating sequence $(\phi_n)_n \subset \Cr$ which converges to $1_E$ in both $L^p(\nu)$ and $L^q(\nu)$.
My attempt
Using mollifiers, we can find a sequence $(\phi_n)_n$ which converges to $1_E$ pointwise almost everywhere, and which converges in every $L^p$ space with the Lebesgue measure. I would then like to use the Dominated Convergence theorem to show that this convergence is in $L^p(\nu)$ and $L^q(\mu)$. However, this would require that $\phi_n \rightarrow 1_E$ pointwise $\mu$- and $\nu$-a.e. We don't know this is true, since $(\phi_n)_n$ may not converge pointwise to $1_E$ on $\partial E$, and we don't know the "behavior" of $\mu$ and $\nu$.
 A: We must assume that $\mu$ and $\nu$ are bounded on bounded sets. Moreover, if $p \neq q$, we must require $E$ to be bounded.
By corollary 4.2.2 of Bogachev's Measure Theory, we know that $C_c^{\infty}(\mathbb{R}^N)$ is dense in $L^{\max(p,q)}(\mathbb{R}^N, \mu+\nu)$. Then we can pick a sequence $(\phi_n)_n \subset C_c^{\infty}(\mathbb{R}^N)$ which converges to $1_E$ in this space. Then it is straightforward to show that for any $f \in L^1(\mathbb{R}^N,\mu+\nu)$, we have:
\begin{align*}
\int |f| \; d(\mu+\nu) = \int |f| \; d\mu + \int |f| \; d\nu \ \geq \int |f| \; d\mu
\end{align*}
From this we conclude that $\phi_n \rightarrow 1_E$ in both $L^{\max(p,q)}(\mathbb{R}^N,\mu)$ and $L^{\max(p,q)}(\mathbb{R}^N,\nu)$. 
If $p = q$, we are done. Otherwise, pick some $R>0$ such that $E \subset B(0,R)$. Then pick a function $v \in C_c^{\infty}(\mathbb{R}^N)$ such that $v = 1$ on $B(0,R)$ and $v=0$ outside of $B(0,2R)$. By inclusions of $L^p$ spaces on sets of finite measure, we can compute that:
\begin{align*}
\int_{\mathbb{R}^N} |v \phi_n - 1_E|^q d\mu
&=
\int_{B(0,R)} |\phi_n - 1_E|^q d\mu + \int_{B(0,2R) \backslash B(0,R)} |v\phi_n|^q d\mu
\\ &\leq
\int_{B(0,2R)} |\phi_n - 1_E|^q d\mu
\\ &\leq
C \int_{B(0,2R)} |\phi_n - 1_E|^{\max(p,q)} d\mu
\\ &\leq
C \int_{\mathbb{R}^N} |\phi_n - 1_E|^{\max(p,q)} d\mu
\end{align*}
Thus we conclude that $v\phi_n \rightarrow 1_E$ in $L^q(\mu)$, and similarly for $L^p(\nu)$.
