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 ?
Best Answer
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.