The following question arises naturally in my current research. It seems to be a basic problem in measure theory, and therefore I guess that answers to it can be found in some textbooks. However, I looked in a few books and found nothing. Since measure theory is not really my field of interest (I do symplectic geometry), I would appreciate any relevant observations and/or references to a text handling this issue.

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$, let $M_\mathbb{C}(\Omega)$ denote the space of complex Borel measures on $\Omega$, and let $M_+(\Omega)$ denote the space of positive Borel measures on $\Omega$. For every complex measure $\lambda\in M_\mathbb{C}(\Omega)$, the total variation of $\lambda$, denoted by $|\lambda|$, is a positive measure. In other words, we have$$|\cdot|:M_\mathbb{C}(\Omega)\to M_+(\Omega).$$We define weak convergence of measures in the usual manner, when thinking of a measure as a linear functional on the space of continuous functions. Namely, we say the sequence $\lambda_1,\lambda_2,\ldots$ of measures converges weakly to the measure $\lambda$, if we have$$\int_\Omega\varphi\lambda_n\to\int_\Omega\varphi\lambda$$for every continuous compactly supported $\varphi:\Omega\to\mathbb{R}$.

Question: Is $|\cdot|$ continuous with respect to weak convergence? That is, if the sequence $(\lambda_n)$ converges weakly to $\lambda$, does it follow that the sequence $(|\lambda_n|)$ converges weakly to $|\lambda|$? If the answer is no, can we add some assumptions on the measures in question to change the picture?

My intuition tells me that the answer should be positive without any further assumptions, but then again, I don't really know...

As written above, any observations are welcome.

  • $\begingroup$ The total variation map $|.|$ is not continuous. A simple example for $\Omega = [0,1]$ is $\lambda_n = \delta_0 - \delta_{\frac{1}{n}}$ for which $\lambda_n \to 0$ but $|\lambda_n| = \delta_0$ for all $n$. An example for which $\lambda_n \to 0$ but $|\lambda_n|$ does not even converge is: $\lambda_n = 0$ if $n$ odd and $\lambda_n = \sin(2\pi n x) dx$ if $n$ even. However, if $\lambda_n$ converges then $|\lambda_n|$ is norm-bounded and thus has at least a convergent subsequence (Helly's selection principle). Moreover, if $\lambda_n(\Omega) \to \lambda(\Omega)$ then $|\lambda_n| \to |\lambda|$. $\endgroup$ – yadaddy Oct 31 '15 at 15:52
  • $\begingroup$ @yadaddy Thanks for your comment. Could you please reference me to a text explaining all that in detail? And about your final statement - it doesn't seem to hold for most of the examples mentioned here (or maybe I'm misinterpreting). Are you sure about it...? Maybe there is another condition to be imposed? $\endgroup$ – Amitai Yuval Nov 1 '15 at 8:17
  • $\begingroup$ A good reference is Bogachev, Measure Theory, Volume II. The last statement is Corollary 8.4.8. It gives a condition (preservation of the total mass in the limit) in addition to $\lambda_n \to \lambda$ which then asserts that the positive and negative parts $\lambda^\pm_n$ converge to the positive resp. negative parts of the limit $\lambda^\pm$. $\endgroup$ – yadaddy Nov 1 '15 at 8:24
  • $\begingroup$ Note also that in a metric space $\Omega$ (as in your case) or more general in a perfectly normal space the Borel $\sigma$-algebra and the Baire $\sigma$-algebra coincide (see Bogachev, Proposition 6.3.4). Thus in this case a Baire measure is the same as a Borel measure (see Definition 7.1.1). $\endgroup$ – yadaddy Nov 1 '15 at 8:38
  • $\begingroup$ Also, in the computations above, some small correction: If $\lambda_n = \delta_0 - \delta_{\frac{1}{n}}$ then $|\lambda_n| = \delta_0 + \delta_{\frac{1}{n}}$ and thus converges to $2 \delta_0$. (This can also be found as Example 8.4.5 in Bogachev). Another good reference for this kind of weak-*convergence (usually referred to as vague convergence) is Bourbaki, Integration. $\endgroup$ – yadaddy Nov 1 '15 at 8:46

This is not in general true. Consider the unit interval in $\mathbb{r}$ and the sequence of complex measures

$$ \lambda_n(A) := \int_A e^{n\pi it} dL $$

where $L$ is Lebesgue measure. The Riemann-Lebesgue lemma says that this converges weakly to the zero measure.

However, we can find a sequence of sets $A_n$, where $A_n \in [0,1]$, such that $\lambda_n(A_n)$ is a non-zero constant (take the set $\{ t: Re(e^{n\pi it}) > 0 \}$). This means that the total variation of $\lambda_n$ cannot converge to the zero measure.

  • $\begingroup$ Thanks for the example! Actually, the total variation of any $\lambda_n$ is just $L$, isn't it so? $\endgroup$ – Amitai Yuval Oct 17 '15 at 7:17
  • $\begingroup$ Yes, you are right. $\endgroup$ – jwg Oct 17 '15 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.