How can we prove that the space of Complex Measures is Complete? with the norm of Total Variation. I have stuck on the last part of the proof where I have to prove that the limit function of a Cauchy sequence of measures has the properties of Complex measures.

We are using the norm of total variation :$$\lVert \mu\rVert = \lvert \mu \rvert(X)$$

  • 1
    $\begingroup$ "... the properties of complex measures." Which ones specifically? Finite additivity should be straightforward if you already showed that there is a set function $\mu$ such that $\mu_n(E) \to \mu(E)$ for all $E \in \Sigma$. To prove $\sigma$-additivity it suffices to show that for every descending sequence $E_{k} \supset E_{k+1}$ of measurable sets such that $\bigcap_{k} E_k = \emptyset$ you have $\mu(E_k) \to 0$. $\endgroup$
    – t.b.
    Aug 4, 2012 at 23:33
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    $\begingroup$ I need just to prove σ-additivity. I find difficulty on that. Can you do it in full detail ? Thank you. $\endgroup$
    – Dimitris
    Aug 5, 2012 at 0:28
  • $\begingroup$ Of course we have to use the total variation, since ∣μ∣(X) is the norm we are using in this space. $\endgroup$
    – Dimitris
    Aug 5, 2012 at 9:58
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    $\begingroup$ @DimitrisNt: To have the total variation norm, you should add $\lVert \mu\rVert = \lvert \mu \rvert(X)$. What you wrote (or copied from the answer) is the total variation measure. $\endgroup$
    – Mirjam
    May 19, 2013 at 11:56
  • $\begingroup$ Yeah , I was studying this proof, and I thought we should refer to the norm we use. I fixed it now $\endgroup$
    – Dimitris
    May 19, 2013 at 11:58

3 Answers 3


As discussed in the comments, the actual question is $\sigma$-additivity of the limit of a Cauchy sequence of complex measures. If you're only interested in this part you can jump to the claim towards the end of the answer, but for the sake of completeness I'll give the definitions and the entire argument that the space of complex measures of bounded variation is a Banach space.

Suppose $(X,\Sigma)$ is a measurable space. A complex measure is a function $\mu\colon \Sigma \to \mathbb C$ satisfying $\mu\left(\bigcup_{k=1}^\infty E_k\right) = \sum_{k=1}^\infty \mu(E_k)$ for every sequence of pairwise disjoint measurable sets $E_k \in \Sigma$.

For a complex measure $\mu$ and $E \in \Sigma$ define $$ \lvert\mu\rvert(E) = \sup\left\{\sum_{k=1}^n \lvert\mu(A_k)\rvert\,:\,A_1,\dots,A_n \subset E\text{ pairwise disjoint and measurable, }n\in\mathbb{N}\right\}. $$ Then the map $\lvert\mu\rvert\colon \Sigma \to [0,\infty]$ is a measure called the total variation of $\mu$ and $\lVert\mu\rVert = \lvert\mu\rvert(X)$ is the total variation norm of $\mu$. If $\lVert\mu\rVert \lt \infty$ then $\mu$ is said to be of bounded variation.

Note that the above definitions also make sense for finitely additive measures, that is to say: functions $\mu\colon \Sigma \to \mathbb{C}$ satisfying $\mu(E \cup F) = \mu(E) + \mu(F)$ whenever $E,F \in \Sigma$ are disjoint.

Let $M(X,\Sigma)$ be the vector space of complex measures of bounded variation and let $M_{\rm fin}(X,\Sigma)$ be the space of finitely additive complex measures of bounded variation, both equipped with the total variation norm. Then $M_{\rm fin}(X,\Sigma)$ is a Banach space and $M(X,\Sigma)$ is a closed subspace, in particular $M(X,\Sigma)$ is a Banach space.

Given a Cauchy sequence $(\mu_n)_{n=1}^\infty$ in $M(X,\Sigma)$, we need to show that there is a measure $\mu$ such that $\lVert\mu - \mu_n\rVert \xrightarrow{n\to\infty}0$. The candidate measure $\mu$ is easily found: For every $E \in \Sigma$ we have that $\rvert\mu_k(E) - \mu_l(E)\rvert \leq \lVert\mu_k-\mu_l\rVert \xrightarrow{k,l\to\infty}0$, so it makes sense to define $$\mu(E) = \lim_{k\to\infty}\mu_k(E) \in \mathbb{C}.\tag{$1$}$$

It follows directly from the definition of $\mu$ as a limit that for disjoint $E,F \in \Sigma$ we have $$ \mu(E \cup F) = \lim_k\mu_k(E\cup F) = \lim_k\mu_k(E) + \lim_k\mu_k(F) = \mu(E)+\mu(F), $$ so $\mu$ is a finitely additive complex measure. We have $$ \lVert\mu-\mu_m\rVert \leq \liminf_{n\to\infty}\lVert\mu_n-\mu_m\rVert \xrightarrow{m\to\infty}0, $$ and similarly $\lVert \mu\rVert \lt \infty$, so in particular we've shown that the space $M_{\rm fin}(X,\Sigma)$ of finitely additive complex measures on $(X,\Sigma)$ with bounded variation is a Banach space and that $\mu$ is the limit of $\mu_n$ with respect to the total variation norm.

Now I finally come to the actual question raised in the comments, namely how to show that the limit $\mu$ of the Cauchy sequence $(\mu_n)$ of complex measures is $\sigma$-additive.

Claim: For every decreasing sequence $F_n \in \Sigma$, that is $F_n \supset F_{n+1}$ for all $n$, such that $\emptyset = \bigcap_{n=1}^\infty F_n$, we have $\mu(F_n) \xrightarrow{n\to\infty}0$.

Note that the claim implies $\sigma$-additivity of $\mu$: Let $E_k \in \Sigma$ be a sequence of pairwise disjoint measurable sets. Put $E = \bigcup_{k=1}^\infty E_k$, let $F_n = \bigcup_{k=n}^\infty E_k = E \smallsetminus \bigcup_{k=1}^{n-1}E_k$ and observe that $\bigcap_{n=1}^\infty F_n = \emptyset$. The claim gives us $$ \left\lvert\mu(E)-\sum_{k=1}^{n-1} \mu(E_k)\right\rvert = \lvert\mu(F_n)\rvert \xrightarrow{n\to\infty} 0, $$ so $\mu\left(\bigcup_{k=1}^\infty E_k\right) = \sum_{k=1}^\infty \mu(E_k)$.

To prove the claim, take $\varepsilon \gt 0$, fix $k$ so large that for $m \geq k$ we have $\lVert\mu_m-\mu_k\rVert \leq \varepsilon/2$ and take $N$ so large that for all $n\geq N$ we have $\lvert\mu_k(F_n)\rvert \leq \varepsilon/2$ (these choices are possible since $(\mu_j)$ is a Cauchy sequence and because $\mu_k$ is a complex measure of bounded variation). For $m \geq k$ we get from $(1)$ that $$ \lvert \mu(F_n)\rvert \xleftarrow{m\to\infty} \lvert\mu_m(F_n)\rvert \leq \lvert\mu_k(F_n)\rvert + \lVert\mu_m-\mu_k\rVert \leq \varepsilon $$ so that $\lvert\mu(F_n)\rvert \leq \varepsilon$ for all $n \geq N$. As $\varepsilon \gt 0$ was arbitrary, the claim is proved.

This shows that $M(X,\Sigma)$ is a closed subspace of the Banach space $M_{\rm fin}(X,\Sigma)$ and we're done.

Added: In fact, the condition of setwise convergence $\mu(E) = \lim_k \mu_k(E)$ for all $E \in \Sigma$ given in $(1)$ is enough to ensure that $\mu$ is $\sigma$-additive, provided all the $\mu_n$'s are.

This is (part of) the Vitali–Hahn–Saks Theorem and this is a bit harder to prove than the above. Various proofs can be found in good measure theory books, e.g. Bogachev, Theorem 4.6.13, page 275, or in this PNAS note by J.K. Brooks. See also this question and answer by Sam L. for an elementary proof.


Another proof, using Radon-Nikodym.

Set $$ \lambda=\sum_{n=1}^\infty \frac{1}{n^2}|\mu_n|. $$ Then clearly, $\lambda$ is a bounded positive measure, and $\mu_n\ll\lambda$, for all $n\in\mathbb N$. Hence, due to Radon-Nikodym, there exist functions $\{f_n\}\in L^1(\lambda)$, such that $d\mu_n=f_n\,d\lambda$, for all $n\in\mathbb N$, and $$ \|\mu_m - \mu_n\|=\int_X|f_m(x)-f_n(x)|\,d\lambda(x)=\|f_m-f_n\|_{L^1(\lambda)} $$ Thus $\{f_n\}$ is a Cauchy sequence, and therefore, convergent to some $f\in L^1(\lambda)$, since $L^1(\lambda)$ is complete. If the measure $\mu$ is defined as $d\mu=f\,d\lambda$, then $$ \|\mu_n-\mu\|=\int_X|f_n(x)-f(x)|\,d\lambda(x)=\|f_n-f\|_{L^1(\lambda)}\to 0. $$

  • $\begingroup$ I'm sorry I know this must be very obvious, but why $\|\mu_m - \mu_n\|=\int_X|f_m(x)-f_n(x)|\,d\lambda(x)$? $\endgroup$
    – user56202
    Jan 25, 2021 at 23:37
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    $\begingroup$ If $\mu\ll\lambda$ and $d\mu=f\,d\lambda$, with $f\in L^1(\lambda)$, then $\|\mu\|=\|f\|_{L^1(\lambda)}$. $\endgroup$ Jan 26, 2021 at 18:03
  • $\begingroup$ @YiorgosS.Smyrlis - What does the notation $L^{1}(\lambda)$ mean? I thought that we put a set in between the brackets, e.g., $L^{1}(\mathbb{R})$? Thank you for your time. $\endgroup$ May 3, 2021 at 1:14
  • $\begingroup$ $L^p(\lambda)$ is the set of $\lambda-$measurable functions on $X$, such that $\int_X |f|^p d\lambda<\infty$. $\endgroup$ May 5, 2021 at 8:22
  • $\begingroup$ It seems obvious but to be fair I don't see if $\vert \vert \mu_m \vert - \vert \mu_n \vert \vert(X) = \int_{X} \vert f_m - f_n \vert \ d \lambda.$ For sure $\vert \vert \mu_m \vert - \vert \mu_n \vert \vert(X) = \int_{X} g \ d \lambda,$ for some nonnegative real function g but is $g = \vert f_m - f_n\vert$ a. e. ? Moreover does $\vert \vert \mu_m \vert - \vert \mu_n \vert \vert(X) \ge \vert \mu_m - \mu_n \vert (X)$. $\endgroup$
    – MI00
    Aug 19 at 17:15

Let me provide an elementary proof.

Let $\mathfrak M$ a $\sigma-$algebra on a set $X$, and $\mathcal X(\mathfrak M)$ the space of complex of complex measures on $\mathfrak M$, which is a equipped with the norm $$ \|\nu\|=|\nu|(X), \quad \nu\in \mathcal X(\mathfrak M), $$ known as total variation of $\nu$.

Let now $\{\mu_n\}$ be a Cauchy sequence on of complex measures $\mathfrak M$.

Step 1. Since for all $E\in\mathfrak M$, $$ |\mu_m(E)-\mu_n(E)|=|(\mu_m-\mu_n)(E)|\le |\mu_m-\mu_n|(E)\le |\mu_m-\mu_n|(X)=\|\mu_m-\mu_n\| $$ then $\{\mu_n(E)\}$ is also Cauchy sequence, and hence convergent, to a complex number. Define $\,\mu: \mathfrak M\to\mathbb C$ as $$ \mu(E)=\lim_{n\to\infty}\mu_n(E). $$ Clearly, $\mu$ is finitely additive since, is $E_1,\ldots, E_k\in\mathfrak M$, mutually disjoint, then
$$ \mu\Big(\bigcup_{j=1}^k E_j\Big)=\lim_{n\to\infty}\mu_n\Big(\bigcup_{j=1}^k E_j\Big) =\lim_{n\to\infty}\sum_{j=1}^k \mu_n(E_j) =\lim_{n\to\infty}\sum_{j=1}^k \mu(E_j). $$

Step 2. Using the fact that, for all $\vartheta,\varphi \in \mathcal X(\mathfrak M)$ $$ |\vartheta+\varphi|(E)\le |\vartheta|(E)|+|\varphi|(E), \quad \text{for all $E\in\mathfrak M$} $$ we obtain that $$ \big| |\mu_m|(E)-|\mu_n|(E)\big|\le |\mu_m-\mu_n|(E)\le|\mu_m-\mu_n|(X)=\|\mu_m-\mu_n\| $$ and hence the sequence $\{|\mu_n|(E)\}$ is also Cauchy and hence convergent. Define $$ \lambda(E)=\lim_{n\to\infty}|\mu_n|(E), \quad E\in\mathfrak M. $$ Clearly $|\mu(E)|\le \lambda(E)$. It can be similarly shown that $\lambda$ is finitely additive, and hence monotonic, i.e. $E\subset F\,\Rightarrow\, \lambda(E)\le \lambda(F)$. Also $\lambda$ is and bounded by $\lambda(X)=\lim_{n\to\infty}|\mu_n|(X)<\infty$.

Step 3. We shall need the following property of $\lambda$. (Which in fact implies that $\lambda$ is a measure.)

If the sets $\{E_j\} \subset \mathfrak M$, are mutually disjoint, and $F_n=\bigcup_{j>n}E_j$, then $\lim_{n\to\infty}\lambda(F_n)=0$.

Proof of the property. Let $E=\bigcup_{j=1}^\infty E_j$. Then $$ \lambda(E)=\lambda \big(\bigcup_{j=1}^n E_j\Big)+\lambda(F_n) $$ and hence the sequence $\{\lambda(F_n)\}$ is decreasing. Let $\varepsilon>0$. Then there exists an $N\in\mathbb N$, such that $$ \big||\mu_m|(G)-|\mu_n|(G)\big|\le\|\mu_m-\mu_n\|<\frac{\varepsilon}{2}, \quad \text{for all $m,n\ge N$ and $G\in\mathfrak M$.} $$ Hence $$ \big| |\mu_n|(G)-\lambda(G)\big|\le \frac{\varepsilon}{2}, \quad \text{for all $n\ge N$ and $G\in\mathfrak M$.} \tag{1} $$ Fix $n_0\ge N$. Clearly, as $|\mu_{n_0}|$ is a bounded positive measure, then $\lim_{n\to\infty}|\mu_{n_0}|(F_n)=0$, and hence, there exists an $N_1\in\mathbb N$, such that $$ |\mu_{n_0}|(F_n)<\frac{\varepsilon}{2}, \quad \text{whenever $n\ge N_1$}. \tag{2} $$ Combining $(1)$ and $(2)$, we obtain that $$ \lambda(F_n)<\varepsilon, \quad \text{whenever $n\ge N_1$,} $$ and hence $\lambda(F_n)\to 0$.

Step 4. $\mu$ is a measure.

Assume that the sets $\{E_j\}\subset\mathfrak M$ are mutually disjoint, $E=\bigcup_{j=1}^\infty E_j$ and $F_n=\bigcup_{j>n}E_j$. Then $\mu(E)=\mu\big(\bigcup_{j=1}^nE_j\big)+\mu(F_n)$ and hence $$ \big|\mu(E)-\sum_{j=1}^n\mu(E_j)\Big|=|\mu(F_n)|\le \lambda(F_n)\to 0, $$ and hence $\sum_{j=1}^\infty \mu(E_j)$ converges and $\mu(E)=\sum_{j=1}^\infty \mu(E_j)$.

Step 5. It remains to show that $\|\mu_n-\mu\|\to 0$.

Let $\varepsilon>0$. Then there exists an $N\in\mathbb N$, such that $$ \|\mu_m-\mu_n\|<\frac{\varepsilon}{4}, \text{whenever $m,n\ge N$}. \tag{3} $$ If $n\ge N$, then there exists a partition $\{E_j\}$ of $X$, such that $$ \|\mu_n-\mu\|<\sum_{j=1}^\infty |(\mu_n-\mu)(E_j)|+\frac{\varepsilon}{4} $$ and for some $k\in\mathbb N$, $$ \|\mu_n-\mu\|<\sum_{j=1}^\infty |(\mu_n-\mu)(E_j)|+\frac{\varepsilon}{4} <\sum_{j=1}^k |(\mu_n-\mu)(E_j)|+\frac{2\varepsilon}{4}. \tag{4} $$ Since $\mu_n(E_j)\to \mu(E_j)$, for all $j=1,\ldots,k$, we can pick $N_1\ge N$, such that $$ \sum_{j=1}^k |(\mu_m-\mu)(E_j)|<\frac{\varepsilon}{4}, \quad\text{whenever $m\ge N_1$} $$ and hence, whenever $m\ge N_1$, \begin{equation} \begin{aligned} \sum_{j=1}^k |(\mu_n-\mu)(E_j)|\le\sum_{j=1}^k |(\mu_n-\mu_m)(E_j)|+ \sum_{j=1}^k |(\mu_m-\mu)(E_j)| \\ <\sum_{j=1}^k |(\mu_m-\mu_n)(E_j)|+ \frac{\varepsilon}{4} \le \|\mu_m-\mu_n\|+ \frac{\varepsilon}{4}. \end{aligned} \tag{5} \end{equation} and hence, combining $(3),\,(4)$ and $(5)$, we obtain that $$ \|\mu_n-\mu\|<\varepsilon, \text{whenever $n\ge N$.} $$


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