# Vector-valued integration with respect to a complex measure

Rudin's Functional Analysis defines the vector-valued integration in Definition 3.26 as:

Suppose $$\mu$$ is a measure on a measure space $$Q$$, $$X$$ is a topological vector space on which $$X^*$$ separates points, and $$f$$ is a function from $$Q$$ into $$X$$ such that the scalar functions $$\Lambda f$$ are integrable with respect to $$\mu$$, for every $$\Lambda\in X^*$$; note that $$\Lambda f$$ is defined by $$\begin{equation} (\Lambda f)(q) = \Lambda(f(q)) \quad (q\in Q). \end{equation}$$ If there exists a vector $$y\in X$$ such that $$\begin{equation} \Lambda y = \int_Q(\Lambda f)d\mu \end{equation}$$ for every $$\Lambda\in X^*$$, then we define $$\begin{equation} \int_Qfd\mu = y. \end{equation}$$

Then, an existence theorem of the vector-valued integration is provided in Theorem 3.27 as:

Suppose (a) $$X$$ is a topological vector space on which $$X^*$$ separates points, and (b) $$\mu$$ is a Borel probability measure on a compact Hausdorff space $$Q$$. If $$f:Q\to X$$ is continuous and if $$\overline{co}(f(Q))$$ is compact in $$X$$, then the integral $$\begin{equation} y = \int_Qfd\mu \end{equation}$$ exists, in the sense of Definition 3.26. Moreover, $$y\in\overline{co}(f(Q))$$.

Remark to Theorem 3.27 says

If $$\nu$$ is any positive Borel measure on $$Q$$, then some scalar multiple of $$\nu$$ is a probability measure. The theorem therefore holds (except for its last sentence) with $$\nu$$ in place of $$\mu$$. It can then be extended to real-valued Borel measures (by the Jordan decomposition theorem) and (if the scalar field of $$x$$ is $$C$$) to complex ones.

When I read Definition 3.26, I understood $$\int_Q(\Lambda f)d\mu$$ as the integration of a complex function $$\Lambda f$$ with respect to a complex measure $$\mu$$, defined in the section 6.18 of Rudin's Real and Complex Analysis as:

If $$\mu$$ is a complex Borel measure, Theorem 6.12 asserts that there is a complex Borel function $$h$$ with $$|h| = 1$$ such that $$d\mu = hd|\mu|$$. It is therefore reasonable to define integration with respect to a complex measure $$\mu$$ by the formula $$\begin{equation} \int fd\mu = \int fhd|\mu|. \end{equation}$$

However, Remark to Theorem 3.27 seems to imply $$\begin{equation*} \int_Q(\Lambda f)d\mu = \left(\int_Q(\Lambda f)d\Re(\mu)^+ - \int_Q(\Lambda f)d\Re(\mu)^-\right) + i\left(\int_Q(\Lambda f)d\Im(\mu)^+ - \int_Q(\Lambda f)d\Im(\mu)^-\right) \end{equation*}$$ where $$\Re(\mu) = \Re(\mu)^+ - \Re(\mu)^-$$ and $$\Im(\mu) = \Im(\mu)^+ - \Im(\mu)^-$$ are the Jordan decompositions of the real and imaginary parts of $$\mu$$.

Since the polar decomposition $$d\mu = hd|\mu|$$ does not guarantee continuity of $$h$$, I guess that Theorem 3.27 cannot be used to prove the existence of $$y\in X$$ satisfying $$\begin{equation} \Lambda y = \int_Q(\Lambda f)hd|\mu| = \int_Q(\Lambda(hf))d|\mu| \end{equation}$$ for all $$\Lambda\in X^*$$ where $$(\Lambda(hf))(q) = \Lambda(h(q)f(q))$$.

So, I am a bit confused with two different definitions of the Lebesgue integration with respect to a complex measure. Are they same? If not, what are the major differences between them and is there any reference that compares them?

• Show that the are the same (for 'ordinary functions') in the usual manner, indicator functions, simple functions and then general functions. Jul 8, 2020 at 4:42
• @copper.hat Thanks for your help. By following your comment, I could prove that they are same at least in $L^1(|\mu|)$. Jul 9, 2020 at 1:36
• You are welcome - this is a repetitive theme with measures & integration (indicator, simple, general). Jul 9, 2020 at 1:38

If $$\mu$$ is a complex measure on a $$\sigma$$-algebra $$\mathfrak{M}$$ in $$X$$, $$d\mu = hd|\mu|$$ is the polar decomposition of $$\mu$$, $$\mu = \Re(\mu) + i\Im(\mu) = \mu_1 - \mu_2 + i(\mu_3 - \mu_4)$$ is the Jordan decompositions of the real and imaginary parts of $$\mu$$, $$f\in L^1(|\mu|)$$, and $$E\in\mathfrak{M}$$, then $$\begin{equation} \int_Efhd|\mu| = \left(\int_Efd\mu_1 - \int_Efd\mu_2\right) + i\left(\int_Efd\mu_3 - \int_Efd\mu_4\right). \tag{1} \end{equation}$$
proof: Since $$|h| = 1$$ and $$\mu_j(A) \le |\mu|(A)$$ for every $$j$$ and $$A\in\mathfrak{M}$$, $$f\in L^1(|\mu|)$$ implies $$fh\in L^1(|\mu|)$$ and $$f\in \bigcap_{j=1}^4L^1(\mu_j)$$. Thus, the left and the right sides of (1) are well defined. Since $$|\mu|$$ is bounded and \begin{align*} \int_E\chi_Ahd|\mu| &= \mu(E\cap A) \\ &= \mu_1(E\cap A) - \mu_2(E\cap A) + i(\mu_3(E\cap A) - \mu_4(E\cap A)) \\ &= \left(\int_E\chi_A\mu_1 - \int_E\chi_A\mu_2\right) + i\left(\int_E\chi_A\mu_3 - \int_E\chi_A\mu_4\right) \end{align*} for every characteristic function $$\chi_A$$ for $$A\in\mathfrak{M}$$, (1) holds for every simple measurable function. Since simple measurable functions are dense in $$L^1(|\mu|)$$, there is a sequence $$\{f_k\}$$ in $$L^1(|\mu|)$$ such that (1) holds for each $$f_k$$ and $$f_k \to f$$ as $$k\to\infty$$ in $$L^1(|\mu|)$$. Since $$L^1(|\mu|) \subset \bigcap_{j=1}^4L^1(\mu_j)$$, $$f_k$$ also converges to $$f$$ in $$\bigcap_{j=1}^4L^1(\mu_j)$$. Therefore, \begin{align*} &\left|\int_Efhd|\mu| - \left(\int_Efd\mu_1 - \int_Efd\mu_2\right) - i\left(\int_Efd\mu_3 - \int_Efd\mu_4\right)\right| \\ &\le \int_E|f - f_k|d|\mu| + \sum_{j=1}^4\int_E|f-f_k|d\mu_j \\ &\to 0 \quad (k\to\infty). \end{align*}