Complex Measures

If $\mu$ is a complex measure on a $\sigma$-algebra $\mathfrak {M}$, and if $E \in \mathfrak {M}$, define

the supremum being taken over all finite partitions $\left\{ E_{i}\right\}$ of $E$. Does it follow that $\lambda = \left|\mu \right|$?

Prove that the example given at the end of Sec. 6.10 has the stated properties.

Prove that the vector space $M(X)$ of all complex regular Borel measures on a locally compact Hausdorff space $X$ is a Banach space if $\left\| \mu \right\| = \left|\mu \right|(X)$.

Suppose $1 \leq p \leq \infty$, and $q$ is the exponent conjugate to $p$. Suppose $\mu$ is a positive $\sigma$-finite measure and $g$ is a measurable function such that $f g \in L^{1}(\mu )$ for every $f \in L^{p}(\mu )$. Prove that then $g \in L^{q}(\mu )$.

Suppose $X$ consists of two points $a$ and $b$; define $\mu (\left\{ a\right\} ) = 1, \mu (\left\{ b\right\} ) = \mu (X) = \infty$, and $\mu (\varnothing ) = 0$. Is it true, for this $\mu$, that $L^{\infty }(\mu )$ is the dual space of $L^{1}(\mu )$?

Suppose $1 < p < \infty$ and prove that $L^{q}(\mu )$ is the dual space of $L^{p}(\mu )$ even if $\mu$ is not $\sigma$-finite. (As usual, $1 / p+1 / q = 1$.)

Suppose $\mu$ is a complex Borel measure on $[0,2 \pi$) (or on the unit circle $T$), and define the Fourier coefficients of $\mu$ by

Assume that $\hat{\mu }(n) \rightarrow 0$ as $n \rightarrow +\infty$ and prove that then $\hat{\mu }(n) \rightarrow 0$ as $n \rightarrow -\infty$. Hint: The assumption also holds with $f d \mu$ in place of $d \mu$ if $f$ is any trigonometric polynomial, hence if $f$ is continuous, hence if $f$ is any bounded Borel function, hence if $d \mu$ is replaced by $d\left|\mu \right|$.

In the terminology of Exercise 7, find all $\mu$ such that $\hat{\mu }$ is periodic, with period $k$.

Suppose that $\left\{ g_{n}\right\}$ is a sequence of positive continuous functions on $I = [0,1]$, that $\mu$ is a positive Borel measure on $I$, and that

(i) $\lim_{n \rightarrow \infty } g_{n}(x) = 0$ a.e. $[m]$,

(ii) $\int_{I} g_{n} d m =1$ for all $n$,

(iii) $\lim_{n \rightarrow \infty } \int_{I} f g_{n} d m = \int_{I} f d \mu$ for every $f \in C(I)$.

Does it follow that $\mu \perp m$?

Let $(X, \mathfrak {M}, \mu )$ be a positive measure space. Call a set $\Phi \subset L^{1}(\mu )$ uniformly integrable if to each $\epsilon > 0$ corresponds a $\delta > 0$ such that

whenever $f \in \Phi$ and $\mu (E) < \delta$.

Suppose $\mu$ is a positive measure on $X, \mu (X) < \infty , f_{n} \in L^{1}(\mu )$ for $n = 1,2,3, \ldots , f_{n}(x) \rightarrow f(x)$ a.e., and there exists $p > 1$ and $C < \infty$ such that $\int_{x}\left|f_{n}\right|^{p} d \mu < C$ for all $n$. Prove that

Hint: $\left\{ f_{n}\right\}$ is uniformly integrable.

Let $\mathfrak {M}$ be the collection of all sets $E$ in the unit interval $[0,1]$ such that either $E$ or its complement is at most countable. Let $\mu$ be the counting measure on this $\sigma$-algebra $\mathfrak {M}$. If $g(x) = x$ for $0 \leq x \leq 1$, show that $g$ is not $\mathfrak {M}$-measurable, although the mapping

makes sense for every $f \in L^{1}(\mu )$ and defines a bounded linear functional on $L^{1}(\mu )$. Thus $\left(L^{1}\right)^{*} \neq L^{\infty }$ in this situation.

Let $L^{\infty } = L^{\infty }(m)$, where $m$ is Lebesgue measure on $I = [0,1]$. Show that there is a bounded linear functional $\Lambda \neq 0$ on $L^{\infty }$ that is 0 on $C(I)$, and that therefore there is no $g \in L^{1}(m)$ that satisfies $\Lambda f = \int_{I} f g d m$ for every $f \in L^{\infty }$. Thus $\left(L^{\infty }\right)^{*} \neq L^{1}$.