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?

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

1 Answer 1


By following the comment of copper.hat, I could prove that they are same.

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*}


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