Reverse type $1-1$ inequality If $\mu \in \mathcal{M}(\mathbb{T})$ is nontrivial and singular with respect to lebesgue measure, then 
$$|\{\theta \in \mathbb{T} : M\mu(\theta) >\lambda\}| \ge \frac C\lambda \|\mu\|$$
where $| \cdot | $ denotes Lebesgue measure, $|\lambda| > \|\mu\|$, which is the total variation norm, and $C$ is an absolute constant.
That is the problem statement. I'm looking for advice on how to establish a weak 1-1 inequality going the other way. The standard tools (Besicovitch, Vitali) don't seem readily applicable. 
The maximal function $M\mu(x) = \displaystyle\sup_{I \supset x} \frac{\mu(I)}{|I|}$ where the sup is taken over all intervals $I$ containing $x$.
 A: I think I have a proof. Assume that $\mu$ is positive and let $\mu = \mu_d + \mu_s$ be the Lebesgue decomposition, where $\mu_d$ is the discrete part and $\mu_s$ the absolutely singular.
Case 1: $\mu_d \neq 0$. The result can be proved as follows. Note that $M(\mu_1) \leq M(\mu_2)$, if $\mu_1 \leq \mu_2$. We have that $$\mu_d = \sum_{j} a_j \, \delta_{\theta_j}$$, choose one $j$, with $a_j > 0$, then $a_j \, M(\delta_{\theta_j}) \leq M(\mu)$. Therefore:
\begin{eqnarray}
  |\{ \theta : M(\mu) > \lambda \}| & \geq & |\{ \theta : a_j \, M(\delta_{\theta_j}) > \lambda \}|\\
  & = & \frac{2 \, a_j}{\lambda}.
\end{eqnarray}
In the second equality we have used that $M(\delta_{\theta_j}) = 1/|\theta -\theta_j|$. 
Case 2: $\mu_d = 0$. This case is much more subtle. We have that $\mu = \mu_s$ s absolutely singular and so its support $S$ is a set of measure zero (without isolated points). By inner regularity we can pick a compact set $K \subset S$ such that $1- \epsilon \leq \mu(K)$. We can work with $d \nu = \mathbf{1}_K \, d \mu$, since $\nu \leq \mu$ and after a normalization we can consider $\nu$ to be a probability measure. Since the support of $\nu$ is a compact we have
$$
  K = \mathbb{T} \setminus \coprod_{j = 1}^\infty I_j,
$$
where $I_j$ are a family of disjoint open intervals. Fix an integer $N > 1$, to be determined latex in terms of $\lambda$. Define
\begin{eqnarray}
  K_N & = & \mathbb{T} \setminus \coprod_{j = 1}^N I_j \\
  L_N & = & |K_N| \rightarrow 0^+.
\end{eqnarray}
Note that $K_N = \coprod_{j=1}^N J_j$, where $J_j$ are closed intervals. Define $q_j = \nu(J_j)$ and $p_j = |J_j|$. We have that
$$
\begin{eqnarray}
  \sum_{j=1}^N p_j & = & |K_N| \\
  \sum_{j=1}^N q_j & = & 1.
\end{eqnarray}
$$
Now, for each $N$ we define a function $\varphi_N: K_N^c \rightarrow \mathbb{R}_+$ such that $(M\nu)(\theta) \geq \varphi_N(\theta)$ given by
$$
  \varphi_N(\theta) = \frac{q_j}{|\theta - b_j| + p_j},
$$
where $\theta \in I_j$ and $J_j = [a_j, b_j]$, $I_j = (a_j, c_j)$ are contiguous intervals. To see that $\varphi_N(\theta) \leq M(\nu)(\theta)$ for $\theta \in K_N^c$ just notice that it is equal to
$$
  \varphi_N(\theta) = \frac{\nu((a_j,\theta))}{|(a_j,\theta)|}.
$$
Now
\begin{eqnarray}
  |\{ \theta \in \mathbb{T} : M(\nu) > \lambda \}| & \geq & |\{ \theta \in K_N^c : M(\nu) > \lambda \}|\\
  & \geq & |\{ \theta \in K_N^c : \varphi_N(\theta) > \lambda \}| \\
  & = & \sum_{j=1}^N \frac{q_j}{\lambda} - p_j = \frac{1}{\lambda} - |K_N|,
\end{eqnarray}
takings $N$ large enough so that $|K_N| <  (\lambda^{-1})/2$ gives the result.
