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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 ?

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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.

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  • $\begingroup$ Thanks, I gave up trying to prove it using what was given in the book up until that point and used the Radon-Nikodym theorem instead, which basically like what you did here. $\endgroup$ – TuoTuo May 13 '15 at 21:40
  • $\begingroup$ You could also consider using the dominated convergence theorem, which is still further in Bass's book than this problem, but less high powered than the Radon-Nikodym theorem. $\endgroup$ – mlg4080 Oct 25 '15 at 3:25
  • $\begingroup$ Can you explain what you mean by the notation $\frac{d\mu}{dm}$? $\endgroup$ – TuringTester69 Nov 25 '18 at 20:05
  • $\begingroup$ @JaneDoe its the Radon-Nikodym derivative. If, for measurable $A$, $\mu(A) > 0$ implies $\nu(A) > 0$, then there is a (unique up to a set of measure zero) function, $f$ st $\int_A f\ d\nu = \mu(A)$ for all measurable $A$, and we write $f = \frac{d\mu}{d\nu}$. Its important to note that in general $f$ does not exist eg the measure on $[0, 1]$ where the points where $1$s occur twice as often as other digits in their decimal expansion doesn't have a derivative wrt Lebesgue measure (look up normal numbers if you're interested in this). $\endgroup$ – user24142 Nov 26 '18 at 18:21
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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.

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  • $\begingroup$ One of the assumptions is that $\mu_{1}(X) < \infty$. Doesn't this counterexample violate that assumption or am I missing something? $\endgroup$ – TuoTuo May 12 '15 at 20:31

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