Show $\mu$ is absly cts w.r.t. $m$ I am working on some practice questions for mt, and am struggling a bit with the last few parts of this question:

Let $\Omega$ be a $\sigma$ of subsets of $X$ and let $\mu$,
  $m$ be finite measures.
$B(X) $
be a v-space of measurable fns that are zero a.e. For $A \in \Omega$
$\mu_a(A) := inf_{\rho \in B(x)}\int_A  and $\mu_t(A) := \mu_a(A)-\mu(A)$

In the first 3 parts of the question, I was able to show the following facts:

(1) $\mu_a(A) \le \mu(A)$
(2) $\mu_a(A \cup B) $
(3) $\mu(A)$ is a measure on $\Omega$

However, I am quite stuck on the last 3 parts of the question:

(4) Show that $\mu_a << m$

For this part, I know I need to try and show that
$$m(A)=0 \implies \mu_a(A) = inf_{\rho \in B(x)}\int_A  = 0$$
However, I am not really sure how to do this, I can't really see how to connect the two. I know that if $m(A)=0$, then the set $A$ must contain $x \in X | \rho(x) \neq 0$, but how can I then show $\mu_a (A) = 0$?

(5) Let $\eta$ be a measure on $\Omega$ such that $\eta \le \mu$ and $
 \eta << m$. Show that $\eta \le \mu_a$.

I assume the result above to try and tackle this question, but still don't really have any idea how to show this.
Collecting the facts I have so far, I have that:
$\eta \le \mu$, $\mu_a \le \mu$, $\eta << m$ and $\mu_a << m$.
I attempted to prove this question by contradiction, assuming that there exists an $A$ such that $ \eta(A) > \mu_a(A)$, and then shuffled the inequalities around. I also tried to use the fact that $\eta << m, \mu_a << m \implies \exists \space g, h$ such that $\eta(A) = \int_A g dm$ and $\mu_a(A) = \int_A h dm$ respectively, but nothing seems to come to me.

(6) Suppose that $\gamma$ is a measure on $\Omega$ such that $\gamma
 \le \mu_a$ . Show $

This last question has me stuck also, with similar strategies from above - I am still at a dead end. 
Any insight/help would be greatly appreciated!
 A: Are you sure the measure $\mu_t$ is well wrote there? I ask because, the way it is written, we can show $\mu_t\leq 0$ which I suspect makes (6) trivial.
Let $\rho=0$. Then $\rho\in B(X)$, so
\begin{equation}
\mu_a(A)\leq\int_A \vert 1-\rho \vert d\mu\leq\mu(A) 
\end{equation}This implies $\mu_t(A)\leq 0$. That is why I will assume you meant $\mu_t(A)=\mu(A)-\mu_a(A)$.

For this part:

(4) Show that $\mu_a\ll m$

Notice this: 
Let $A$ such that $m(A)=0$. Let $\rho(x)=1_A$ the indicator function in $A$. As $m(A)=0$ then $\rho\in B(X)$. Then 
\begin{equation}
\int_A \vert 1-\rho \vert d\mu=0
\end{equation}which implies $\mu_a(A)=0$
For this part:

(5) Let $\eta$ be a measure on $\Omega$ such that $\eta \le \mu$ and $
 \eta \ll m$. Show that $\eta \leq \mu_a$.

If $\eta(A)=0$, obviously $0=\eta(A)\leq \mu_a(A)$.
If $\eta(A)>0$. Consider $\rho \in B(X)$, then 
\begin{eqnarray}
\mu_a(A)& \leq & \int_A\vert 1-\rho \vert d\mu\\
& = & \int_{A\cap\{\rho\neq 0\}}\vert 1-\rho \vert d\mu+\int_{A\cap\{\rho=0\}}\vert 1-\rho \vert d\mu\\
& = & \mu(A\cap\{\rho=0\})+\int_{A\cap\{\rho\neq 0\}}\vert 1-\rho \vert d\mu
\end{eqnarray}
On the other hand, 
\begin{equation}
\eta(A)=\eta(A\cap\{\rho\neq 0\})+\eta(A\cap\{\rho=0\})
\end{equation}Since $\eta\ll m$ and $\rho\in B(X)\Rightarrow m(A\cap\{\rho\neq 0\})=0\Rightarrow \eta(A\cap\{\rho\neq 0\})=0$. Now, using $\eta\leq \mu$ we have
\begin{equation}
\eta(A)=\eta(A\cap\{\rho=0\})\leq \mu(A\cap\{\rho=0\})\leq \mu(A\cap\{\rho=0\})+\int_{A\cap\{\rho\neq 0\}}\vert 1-\rho \vert d\mu
\end{equation}for all $\rho\Rightarrow \eta(A)\leq\mu_a(A)$
For this part:

(6) Suppose that $\gamma$ is a measure on $\Omega$ such that $\gamma
 \le \mu_a$ and $\gamma \le \mu_t$. Show that $\gamma = 0$

Since I assume $\mu_t=\mu-\mu_a$, then
\begin{eqnarray}
\gamma & \leq & \mu-\mu_a\\
\gamma & \leq & \mu_a
\end{eqnarray}
which implies $2\gamma\leq\mu$. Note $\gamma\leq\mu_a\ll m\Rightarrow \gamma\ll m\Rightarrow 2\gamma \ll m$. Using (5) we see $2\gamma \leq\mu_a$. Using this we have
\begin{eqnarray}
\gamma & \leq & \mu-\mu_a\\
2\gamma & \leq & \mu_a
\end{eqnarray}
which implies $3\gamma\leq\mu$. Inductively we can show $n\gamma\leq \mu$ for all $n$, and since $\mu $ is finite this implies $\gamma=0$.
