There is no translation-invariant measure on an infinite-dimensional space

One difficulty in carrying out analysis on an infinite-dimensional normed vector space, e.g. $L^p(\mathbb{R})$, is that there is no analogue of Lebesgue measure on such a space. This comes from the observation that an open ball in an infinite-dimensional normed vector space contains infinitely many disjoint open balls of the same radius. We first give a formal statement of our goal.

Theorem. Let $X$ be an infinite-dimensional normed vector space. Then there is no translational-invariant Borel measure $\mu$ on $X$ such that $\mu(U) > 0$ for every nonempty open set $U$ and $\mu(U_0) < \infty$ for some open set $U_0$.

To prove the theorem, we recall Riesz's Lemma.

Lemma (Riesz). Let $X$ be a normed vector space and $Y$ be a proper closed subspace. For every $\epsilon > 0$, there exists $x \in X$ such that $\|x\| = 1$ and $\mathrm{dist}(x, Y) = \inf\{\|x - y\|: y \in Y\} \geq 1 - \epsilon$.

Proof of Theorem: Suppose $\mu$ is a translation-invariant Borel measure on $X$ with $\mu(U) > 0$ for every nonempty open $U \subseteq X$. Fix a nonempty open set $U$ in $X$. Then there exists $r > 0$ and $x_0 \in X$ such that $W = r(U + x_0)$ contains $B = \{x \in X: \|x\| < 2 \}$. Using Riesz's Lemma and induction, we can find a sequence $\{x_n\}$ of unit vectors in $X$ such that $\mathrm{dist}(x_{n + 1}, \mathrm{span}\{x_1, \ldots, x_n\}) \geq 1/2$ for all $n \geq 1$. Hence $\{x_n\} \subseteq B$ and $\|x_n - x_m\| \geq 1/2$ whenever $n \neq m$. Take $B_n = \{x \in X: \|x - x_n\| < 1/2 \}$, then $\{B_n\}$ is a disjoint collection of open balls contained in $B$ of radius half. Now let $U_n = \frac{1}{r}B_n - x_0$, then $\{U_n\}$ is a disjoint collection of open balls contained in $U$ of the same radius. It follows that \[ \mu(U) \geq \mu(\bigcup_{n = 1}^\infty U_n) = \sum_{n = 1}^\infty \mu(U_n) = \sum_{n = 1}^\infty \mu(U_1) \] by translation-invariance of $\mu$. But $\mu(U_1) > 0$, forcing $\mu(U) = \infty$.

Reference:
Fabian, Habala, Hajek, Montesinos, Zizler - Banach Space Theory: The Basis for Linear and Nonlinear Analysis

A criterion for writing a functional as a linear combination

Let $X$ be a vector space over $\mathbb{F}$ with dual space $X'$ and fix $\phi_1, \ldots, \phi_n \in X'$. We have a useful criterion for when a linear functional $f \in X'$ lies in the span of $\phi_1, \ldots, \phi_n$.

Theorem 1. Let $f \in X'$. Then $f \in \mathrm{span}\{\phi_1, \ldots, \phi_n\}$ if and only if $\cap_{i = 1}^n \ker(\phi_i) \subseteq \ker(f)$.

To prove this, we first establish a general result on when a linear map can factor through another linear map.

Theorem 2. Suppose we are given linear maps $T: X \rightarrow Y$ and $S: X \rightarrow Z$. Then there exists a linear map $R: Z \rightarrow Y$ such that $T = R \circ S$ if and only if $\ker(S) \subseteq \ker(T)$.

Proof: (=>) Clear. (<=) Suppose $\ker(S) \subseteq \ker(T)$. Define $R': \mathrm{im}(S) \rightarrow Y$ by $R'(S(x)) = T(x)$. If $S(x) = S(y)$, then $x - y \in \ker(S)$, so $x - y \in \ker(T)$, i.e. $T(x) = T(y)$. Hence $R'$ is well-defined. It is easy to check that $R'$ is linear. Now let $P$ be the projection of $Z$ onto $\mathrm{im}(S)$ and take $R = R' \circ P$. Then for all $x \in X$, $R \circ S(x) = R'(P(S(x))) = R'(S(x)) = T(x)$. 

Proof of Theorem 1: (=>) Clear. (<=) Take $T: X \rightarrow \mathbb{F}$ and $S: X \rightarrow \mathbb{F}^n$ defined by $T(x) = f(x)$ and $S(x) = (\phi_1(x), \ldots, \phi_n(x))$. Then $\ker(S) = \cap_{i = 1}^n \ker(\phi_i) \subseteq \ker(f) = \ker(T)$. By Theorem 2 there exists $R: \mathbb{F}^n \rightarrow \mathbb{F}$ such that $T = R \circ S$ or $f(x) = R(\phi_1(x), \ldots, \phi_n(x))$ for all $x \in X$. Writing $R(y_1, \cdots, y_n) = c_1 y_1 + \cdots + c_n y_n$, we see that $f = c_1 \phi_1 + \cdots + c_n \phi_n$.

Reference:
Holmes - Geometric Functional Analysis and its Applications