# Why the space of complex measures is Banach?

I've read the proof from here: Space of Complex Measures is Banach (proof?) and understood the part that proves that constructed limit is complex measure. But the first part is a bit unclear for me. I don't see why $$\|\mu-\mu_n\| \le \liminf\limits_{m\to\infty} \|\mu_m-\mu_n\|$$ Could anyone help me with that?

## 2 Answers

It can be justified via Fatou's lemma with varying measures, which can be found here: https://en.wikipedia.org/wiki/Fatou%27s_lemma#Fatou.27s_Lemma_with_Varying_Measures. Express the norm as an integral of an indicator function and apply the lemma to the sequence $\nu_n = \mu_n - \mu_m$.

I think there are easier ways to prove this space is complete. Recall a normed space $X$ is a Banach space iff for all $(v_n)_{n=1}^\infty \subset X$ with $$\sum_{n=1}^\infty ||v_n||_X<\infty$$ we have $$\sum_{n=1}^\infty v_n$$ converges in $X$. This should be a straightforward computation.

There's a slick way to prove completeness with the Radon-Nikodym theorem: let $(\mu_n)_{n=1}^\infty$ be a cauchy sequence of complex measures (which is necessarily bounded). Consider the measure $$\nu = \sum_{n=1}^\infty 2^{-n}|\mu_n|$$ Then each $\mu_n$ is absolutely continuous with respect to $\nu$ and by Radon-Nikodym there is a sequence $(h_n)_{n=1}^\infty \subset L^1(\nu)$ such that $\mu_n = h_n \cdot \nu$ for all $n$. Then show $||h_n \cdot \nu||$ in measure norm is equal to $||h_n||_{L^1(\nu)}$ and exploit completeness of $L^1$...

For any $$n\in \mathbb{N}$$ we will prove given inequality.

Now let $$n\in \mathbb{N}$$ and let $$a=\|\mu-\mu_n\|$$. Now, given any $$\delta>0$$ there exists a finite measurable partition $$\{E_i\}_{i=1}^N$$ of $$X$$ s.t. $$a-\delta < \sum_{i=1}^N|\mu(E_i)-\mu_n(E_i)|$$ Now since $$\lim_{m\to\infty} \sum_{i=1}^N|\mu_m(E_i)-\mu_n(E_i)|=\sum_{i=1}^N|\mu(E_i)-\mu_n(E_i)|$$ (notice that we have a finite sum so this holds) there exists $$r\in \mathbb{N}$$ s.t. $$m\geq r$$ implies $$a-\delta < \sum_{i=1}^N|\mu_m(E_i)-\mu_n(E_i)|$$ Taking the supremum over all finite measurable partitions $$\{E_i\}_{i=1}^N$$ of $$X$$ gives $$\|\mu_n- \mu_n\|>a-\delta$$ for $$m\geq r$$. Now it is clear by definition of $$\liminf$$ that $$a-\delta \le \liminf_{m\to\infty} \|\mu_m-\mu_n\|$$. Since $$\delta > 0$$ was arbitrary we are done.