The next question I found in my teacher's notes that are in preparation, so that is not trusted:
Let $\lambda_{1}$, $\lambda_{1}$ and $\mu$ signed measures such that $\lambda_{1}\bot\mu$ y $\lambda_{2}\bot\mu$, then $\left(\lambda_{1}+\lambda_{2}\right)\bot\mu$.
Where, considering $\left(X,\mathcal{M}\right)$ be a measurable space, $\lambda:\mathcal{M}\rightarrow\mathbb{R}$ is called signed measure ( or charge) if:
- $\lambda\left(\phi\right)=0,$
- If $\left\{ E_{i}\right\} _{i=1}^{\infty}$ is a disjoint family of elements in $\mathcal{M}$, then $$\lambda\left(\bigcup_{i=1}^{\infty}E_{i}\right)=\sum_{i=1}^{\infty}\lambda\left(E_{i}\right),$$ where the serie on the right-hand side must be unconditionally convergent.
Moreover, we say that two measures $\lambda$ and $\mu$ are mutually singular if there are disjoint sets $A$, $B$ in $\mathcal{M}$ such that $X=A\cup B$ and $\lambda\left(A\right)=\mu\left(B\right)=0$. In this case we write $\lambda\bot\mu$.
If $\lambda$ is signed measure and $\mu$ is measure, we e say that $\lambda$ and $\mu$ are mutually singular if $\left|\lambda\right|\perp\mu$, where $\left|\lambda\right|$ is the total variation of $\lambda$, i,e, $\left|\lambda\right|=\lambda^{+}+\lambda^{-}$.
Question: Notice that the definition of mutually singular for $\lambda$ and $\mu$ involves the need always $\mu$ is a measure, this is one of the reasons I think the problem is ill-posed, however, I need an example of three charges to show that the problem statement It's not valid.
For a signed measure, we can always write it as a difference of two positive measure by Hahn decomposition theorem so I think the problem is not ill-posed because you can $\lambda$ can be written as $\lambda_1-\lambda_2$ and $\mu=\mu_1-\mu_2$ where $\lambda_i,\mu_j$ are positive measure. Then $\lambda\perp \mu$ just means $\lambda_i\perp \mu_j$ for all $i,j$.