Measurability and Measure

Let $S \subset [0,1]$ be the set defined by the property that $x \in S$ if and only if in the decimal representation of $x$ the first appearance of the digit $2$ precedes the first appearance of the digit $3$. Prove that $S$ is Lebesgue measurable and find its measure.

Define $S^1 = \mathbb {R} / \mathbb {Z}$ endowed with the natural Lebesgue measure. Consider on $S^1$ the equivalence relation $[x] \sim [y] \iff x - y \in \mathbb {Q}$, for $x, y \in \mathbb {R}$ and $[\cdot ]$ denoting the class in $S^1$. For each $\xi \in S^1/\! \sim$, choose a representative $\bar{\xi } \in S^1$, and let $E \subset S^1$ be the set of these points, i.e.,

Show that $E$ is not measurable.

Let $X$ be a set and $\mathcal{D} \subset \mathcal{P}(X)$, $\mathcal{D}$ closed under finite intersection. Denote by $\mathcal{R}$ the ring generated by $\mathcal{D}$. Furthermore, let $\pi$ be the smallest system, $\mathcal{D} \subset \pi \subset \mathcal{P}(X)$ such that $\pi$ is closed under the following operations:

1. finite disjoint unions;

2. differences $A \setminus B$, $B \subset A$. Prove that $\pi = \mathcal{R}$.

Let $\mu^*$ be the Lebesgue outer measure on $\mathbb {R}$ and $A, B$ subsets of $\mathbb {R}$ such that

Prove or disprove that

For $f : [0,1] \to \mathbb {R}$, let $E \subset \{ x \mid f'(x) \text{ exists}\}$. If $m(E) = 0$, show that $m(f(E)) = 0$.

Suppose $A \subset \mathbb {R}$ is Lebesgue measurable and assume that

for any $a, b \in \mathbb {R}$, $a < b$. Prove that $m(A) = 0$.

Choose $0 < \lambda < 1$ and construct the Cantor set $K_\lambda$ as follows: remove from $[0,1]$ its middle part of length $\lambda$; we are left with two intervals $I_1$ and $I_2$. Remove from each of them their middle parts of lengths $\lambda \left|I_1\right|$, $\lambda \left|I_2\right|$, etc.and keep doing this ad infinitum. We are left with the set $K_\lambda$. Prove that the set $K_\lambda$ has Lebesgue measure zero.

Suppose $\mu$ is a positive Borel measure on $\mathbb {R}$ such that (i) $\mu ([0,1]) = 1$, and (ii) $\mu (E) = \mu (x+E)$ for any Borel set $E$ of $\mathbb {R}$ and every $x \in \mathbb {R}$. Does this imply that $\mu$ is the Lebesgue measure? Justify your answer.

Let $\mathcal{U}$ be a $\sigma$-algebra of subsets of a set $X$ and $\mu_n : \mathcal{U} \to \mathbb {R}$ be signed measures such that $\mu (E) = \lim_{n \to \infty } \mu_n(E)$ exists for every $E \in \mathcal{U}$. Prove that $\mu$ is a signed measure.

Let $\lambda$ be Lebesgue measure on $\mathbb {R}$. Show that for any Lebesgue measurable set $E \subset \mathbb {R}$ with $\lambda (E) = 1$, there is a Lebesgue measurable set $A \subset E$ with $\lambda (A) = \frac{1}{2}$.

Let $\mu$ be a complex Borel measure on $[0, \infty )$. Show that if

for all $n = 0, 1, 2, \ldots$, then $\mu = 0$.

Let $A \subset [0,1]$ be a measurable set of positive measure. Show that there exist two points $x' \neq x''$ in $A$ with $x' - x''$ rational.

Let $E = \left\{ \sum_{n=1}^{\infty } \frac{x_n}{3^n} \mid x_n = 0 \text{ or } 1\right\}$. Prove that $E$ is Lebesgue measurable with Lebesgue measure $0$. Also show that $E + E = [0,1]$.

Let $f : \mathbb {R} \to \mathbb {R}^+$ be measurable, and let $\varepsilon > 0$. Show that there exists $g : \mathbb {R} \to \mathbb {R}^+$ measurable such that (i) $\left\| f-g\right\|_{\infty } < \varepsilon$, (ii) for every $r \in \mathbb {R}$, $\left|\{ x \mid g(x) = r\} \right| = 0$.

Let $f$ be the function on $[0,1]$ defined as follows: $f(x) = 0$ if $x$ is a point on the Cantor ternary set and $f(x) = 3^{-p}$ if $x$ is in one of the complementary intervals of length $3^{-p}$.

Let $E$ be a Lebesgue measurable subset of $\mathbb {R}$ with $m(E) < \infty$ and let

Show that

Let $A \subset \mathbb {R}$ be a set of positive Lebesgue measure. Prove that

is continuous at $x = 1$. Use this result to prove that there exists an $\varepsilon > 0$ such that for any $m \in \mathbb {R}$ with $\left|m - 1\right| < \varepsilon$, the line $y = mx$ has a non-void intersection with $A \times A$.

Let $\mu$ be a countably additive measure on a set $S$ with $\mu (S) < +\infty$ that is without atoms, i.e., if $A$ is a measurable set with $\mu (A) > 0$, then there is a measurable set $B \subset A$ such that $0 < \mu (B) < \mu (A)$. Prove that the range of $\mu$ is the closed interval $[0, \mu (S)]$.

Let $X$ be a compact Hausdorff space. Consider the $\sigma$-algebra $\mathcal{B}$ generated by the compact $G_\delta$ sets. Show that any positive measure $\mu$ on $\mathcal{B}$ which is finite on compact sets is automatically regular.

Suppose that $\mu$ is a non-negative Borel measure on $\mathbb {R}^n$. Let

Assume $f(r)$ is finite for all $r > 0$ and assume

Prove that $\mu$ is identically $0$.

Let $\mu$ be a finite Borel measure defined on $\mathbb {R}^n$, and define a function $f$ on $\mathbb {R}^n$ by $f(x) = \mu (B(x,1))$ where $B(x,1)$ denotes the open ball centered at $x$ of radius $1$. Prove that $f$ attains its minimum on each compact set of $\mathbb {R}^n$.

Let $E$ be the set of all numbers in $[0,1]$ which can be written in a decimal expansion with no sevens appearing. Thus,

Let $f : \mathbb {R} \to \mathbb {R}^n$ be a function such that for all $x, y \in \mathbb {R}$

Show that if $E \subset \mathbb {R}$ is a measurable set with $m_1(E) = 0$ then $m_n(f(E)) = 0$.

Let $f : \mathbb {R}^n \to \mathbb {R}$ be an arbitrary function having the property that for each $\varepsilon > 0$, there is an open set $U$ with $\lambda (U) < \varepsilon$ such that $f$ is continuous on $\mathbb {R}^n \setminus U$ (in the relative topology). Prove that $f$ is measurable.

Let $A \subset \mathbb {R}$ be a Lebesgue measurable set and let $rA = \{ rx \mid x \in A\}$ where $r$ is a real number. Assume that $rA = A$ for every non-zero rational number $r$. Prove that either $A$ or $\mathbb {R} \setminus A$ has Lebesgue measure zero.

Let $\mu$ be a $\sigma$-finite measure on the measure space $(X, \mathfrak {m})$. Prove that there exists a probability measure $\nu$ on $(X, \mathfrak {m})$ ($\nu (X) = 1$) such that $\mu$ is absolutely continuous with respect to $\nu$, and $\nu$ is absolutely continuous with respect to $\mu$.

Give an example of first category subset of $[0,1]$ which is of Lebesgue measure $1$.

Let $(X, \mathfrak {X}, \mu )$ be a measure space and suppose $(f_n : X \to \mathbb {C})$ is a sequence of measurable functions with the property that for all $n \geq 1$ and all $\lambda > 0$,

where $c$ is a constant independent of $n$. Let $n_k = 2^k$. Prove that

Let $f \in L^1[0,1]$ and let $g(x) = \int_0^x f(t) \, dt$. If $E$ is a measurable subset of $[0,1]$, show that (i) $g(E) = \{ y : \exists x \in E \text{ with } y = g(x)\}$ is measurable, (ii) $m(g(E)) \leq \int_E \left|f(t)\right| \, dt$.

Given a function $f : \mathbb {R} \to \mathbb {Z}$, show that $\{ x : f \text{ is not continuous at } x\}$ is a Borel set.

Let $A$ and $B$ be (not necessarily Lebesgue measurable) subsets of $\mathbb {R}$ and let $\left|\cdot \right|_e$ stand for Lebesgue outer measure. Prove that if $\left|A\right|_e = 1$ and $\left|B\right|_e = 1$ and $\left|A \cup B\right|_e = 2$ then $\left|A \cap B\right|_e = 0$.

Let $f$ be a bounded Lebesgue measurable function on $\mathbb {R}$. Put

where $\left|\cdot \right|$ is the Lebesgue measure (i.e., $g(x)$ equals the essential supremum of $f$ over $(x, x+1)$). Prove $\varliminf_{x \to 0} g(x) \geq g(0)$.

Show that if $E \subset \mathbb {R}$ and if $\left|\cdot \right|_e$ stands for outer measure then

Also prove that if $O_i$, $i \geq 1$, are open subsets of $\mathbb {R}$ satisfying $\bigcup_{i=1}^{\infty } O_i = \mathbb {R}$ then

If $\nu$ and $\mu$ are measures on a measurable space $(X, \mathcal{M})$, we say that $\nu \leq \mu$ if $\nu (A) \leq \mu (A)$ for all $A \in \mathcal{M}$. Assume that $(\nu_n)$ is a sequence of measures on $(X, \mathcal{M})$ such that $\nu_1 \leq \nu_2 \leq \cdots$. Define $\nu (A) = \lim_{n \to \infty } \nu_n(A)$, $A \in \mathcal{M}$.

Define $f : [0,1] \to \mathbb {R}$ as follows: if $x \in [0,1]$ has decimal expansion $x = 0.a_1a_2a_3\cdots$ then $f(x) = \max_k a_k$. Prove that the function $f$ is measurable.

Let $m$ and $\mu$ be, respectively, the Lebesgue measure and the counting measure on the Borel $\sigma$-algebra on $[0,1]$.