Decreasing Sequence of Measures For $(X,\mathcal{F})$ a measure space, I know that if we have $\mu_{n}(A) \searrow$, i.e. is a decreasing sequence of measures for each $A \in \mathcal{F}$ and $\mu_{1}(X) < \infty$ then $\mu = \lim_{n \rightarrow \infty} \mu_{n}$ is not a measure.
The problem asks for a counterexample. I am struggling with coming up with a counterexample for this and would appreciate some help. This problem comes from an early chapter in the book before any discussion of Lebesgue measure, so it should be possible to come up with a counterexample using only the measures discussed at that point which are the counting measure or Dirac measure and linear combinations of measures.
A hint would be appreciated. 
Edit: I am starting to think that this may actually be a measure if $\mu_{1}(X) < \infty$. Is this the case ?
 A: Let $\mu_0$ assign measure 1 to each point of $\mathbb{N}$. Let $\mu_n$ assign measure 1 to $n,n+1,\ldots$ and zero elsewhere. Clearly $\mu_n$ is decreasing. On the other hand, notice that $\mu_n(\mathbb{N})=\infty$ which implies $\mu(\mathbb{N})=\infty$, yet $\lim_{n\rightarrow\infty}\mu_n(\{k\})=0$ for each $k\in\mathbb{N}$ which implies $\mu(\{k\})=0$. This will break countable additivity. 
Contrast this with an increasing family of measures, where you can use the monotone convergence theorem to verify countable additivity. 
A: I believe that restricting to finite measures forces $\mu$ to be a measure. I can't prove it with what you already know, and I suspect that Alex R has hit on the precise counter example that was expected. But anyway...
Let $m = \mu_1$, and $f_n = \frac{d \mu_n}{d m}$. The sequence $\{f_n\}$ is integrable, monotone decreasing, and bounded below by $0$, so it converges pointwise to a function $f$. We may define a measure $\mu$ to be such that $\mu(A) = \int_A f dm$. By the Lebesgue monotone convergence theorem, for any measurable $A$, $$\lim_{n\to \infty} \mu_n(A) = \lim_{n\to \infty}\int_A f_n dm = \int_A f dm = \mu(A)$$ which is as required.
We use the finiteness of the measures when we say that the $f_n$ are integrable. Without that, you can get the sort of pathological cases, like those illustrated by Alex R.
