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The Lebesgue measure on $\mathbb{R}$ is the same as the Lebesgue measure on $\mathbb{R}$ scaled by a constant $c$.
If $r \neq 0$, then $y = x/r$ if and only if $r \cdot y = x$.
If $f$ is a Bochner integrable function on $\mathbb{R}$, then $f(t + cx)$ is Bochner integrable on $\mathbb{R}$ and $\int_{\mathbb{R}} f(t + cx) \, dx = \frac{1}{|c|} \int_{\mathbb{R}} f(x) \, dx$.
The distribution of $-X$ is the same as the distribution of $X$.
If $c \neq 0$, then the distribution of $cX$ is the same as the distribution of $X/|c|$.
If $c \neq 0$, then the distribution of $cX$ is the same as the distribution of $X$ scaled by $\lvert c \rvert$.
The distribution of the random variable $X + c$ is the same as the distribution of $X$.
The product of the Lebesgue measure on $\mathbb{R}^n$ with itself is the Lebesgue measure on $\mathbb{R}^{2n}$.
If $P$ is a Borel-measurable predicate, then it is also Lebesgue-measurable.
If $A$ is a Borel set, then $A$ is a Lebesgue-Borel set.
If $A$ is a bounded set, then the Lebesgue measure of $A$ is finite.
If $A$ is a compact set, then the Lebesgue measure of $A$ is finite.
If $f$ is a continuous function defined on a compact set $S$, then $f$ is Borel integrable on $S$.
If $f$ is continuous on the interval $[a,b]$, then $f$ is integrable on $[a,b]$.
The whole space is not Lebesgue measurable.
A set $S$ is Lebesgue measurable if and only if the indicator function of $S$ is integrable.
The closed and open boxes are Lebesgue measurable.
The closed interval $[a,b]$ and the open interval $(a,b)$ are Lebesgue measurable.
If $S$ is a compact set, then $S$ is a measurable set.
Any compact set is Lebesgue measurable.
If $S$ is a bounded set, then the measure of the boundary of $S$ is equal to the measure of the closure of $S$ minus the measure of the interior of $S$.
If $S$ is a bounded set, then the closure of $S$ is Lebesgue measurable.
If $S$ is a bounded set, then the frontier of $S$ is Lebesgue measurable.
If $S$ is a bounded open set, then $S$ is Lebesgue measurable.
The ball $B(a,r)$ is Lebesgue measurable.
The closed ball of radius $r$ centered at $a$ is Lebesgue measurable.
If $S$ is a bounded set, then the interior of $S$ is Lebesgue measurable.
The difference between a closed box and an open box is a null set.
If $S$ is a bounded Lebesgue measurable set, then $S$ is Lebesgue measurable.
If $S$ is Lebesgue measurable, then the Lebesgue measure on $S$ is a finite measure.
The constant function $c$ is integrable on the interval $[a,b]$.
If $S$ is a $\sigma$-set, then $f(S)$ is also a $\sigma$-set.
If $N$ is a null set, then the set $\{x - a : x \in N\}$ is also a null set.
If $S$ is a Lebesgue measurable set, then the set $a + S$ is also Lebesgue measurable.
If $S$ is a Lebesgue measurable set, then $S + a$ is also Lebesgue measurable.
If $S$ is a Lebesgue measurable set, then the set $S - a$ is also Lebesgue measurable.
The Lebesgue measure of a translation of a set is equal to the Lebesgue measure of the set.
The Lebesgue measure of a set $S$ is equal to the Lebesgue measure of the set $\{x - a \mid x \in S\}$.
If $f$ is a non-negative function that is not summable, then $\sum_{i=0}^\infty f(i) = \infty$.
If $S$ is a bounded set of Lebesgue measure zero such that for all $c \geq 0$ and $x \in S$, if $cx \in S$, then $c = 1$, then $S$ is a null set.
If $S$ is a compact set such that for all $c \geq 0$ and $x \in S$, if $cx \in S$, then $c = 1$, then $S$ is a null set.
For any Borel set $B$ and any $\epsilon > 0$, there exists an open set $U$ such that $B \subseteq U$ and $\mu(U - B) \leq \epsilon$.
For any Borel set $B$ and any $\epsilon > 0$, there exists an open set $U$ such that $B \subseteq U$ and $\mu(U - B) < \epsilon$.
If $A$ is a measurable set in the completion of a measure space $M$, then there exists a measurable set $A'$ in $M$ such that $A \subseteq A'$, $A' - A$ is a null set in the completion of $M$, and the measure of $A$ in the completion of $M$ is equal to the measure of $A'$ in $M$.
If $S$ is a Lebesgue measurable set, then for every $\epsilon > 0$, there exists an open set $T$ such that $S \subseteq T$, $T - S$ is Lebesgue measurable, and $\mu(T - S) < \epsilon$.
Suppose $S$ is a Lebesgue measurable set. For every $\epsilon > 0$, there exists a closed set $T$ such that $T \subseteq S$, $S - T$ is Lebesgue measurable, and $\mu(S - T) < \epsilon$.
If $T$ is an open subset of a Lebesgue measurable set $S$, then $T$ is Lebesgue measurable.
If $T$ is a closed subset of a Lebesgue measurable set $S$, then $T$ is Lebesgue measurable.
A set $S$ is an $F_\sigma$ set if and only if there exists a sequence of compact sets $F_n$ such that $S = \bigcup_n F_n$.
If $S$ is a $G_{\delta}$ set, then $-S$ is an $F_{\sigma}$ set.
If $S$ is an $F_\sigma$ set, then $-S$ is a $G_\delta$ set.
A set is a $G_\delta$ set if and only if its complement is an $F_\sigma$ set.
If $S$ is a Lebesgue measurable set, then there exists a countable union of closed sets $C$ and a set $T$ of measure zero such that $S = C \cup T$ and $C \cap T = \emptyset$.
Every Lebesgue measurable set is almost a $G_\delta$ set.
A predicate $P$ holds eventually at infinity if and only if there exists a real number $b$ such that $P$ holds for all $x$ with $\|x\| \geq b$.
A predicate $p$ holds eventually at infinity if and only if there exists a positive number $b$ such that $p$ holds for all $x$ with $\|x\| \geq b$.
The filter at infinity is equal to the supremum of the filters at the top and at the bottom.
The filter of events that happen at infinity is at least as strong as the filter of events that happen at the top.
The filter at $\bot$ is a subset of the filter at $\infty$.
If $f$ tends to infinity, then $f$ tends to infinity along any filter.
The filter of real numbers that tend to infinity is the filter of sequences that tend to infinity.
If $f$ converges to $l$ at infinity, then the sequence $(f(n))$ converges to $l$.
A sequence $x_n$ is bounded if and only if the function $n \mapsto x_n$ is bounded.
If $X$ is a bounded sequence, then so is the sequence $X$ shifted by $k$ positions.
If $X_n$ is a bounded sequence, then so is $X_{n+k}$.
A function $f$ is bounded on a filter $F$ if and only if there exists a constant $K > 0$ such that $|f(x)| \leq K$ for all $x$ in the filter $F$.
If $f$ is a function such that $|f(x)| \leq K$ for all $x$ in some set $F$, then $f$ is a bounded function.
If $f$ is a bounded function on a filter $F$, then there exists a positive real number $B$ such that $|f(x)| \leq B$ for all $x$ in the filter.
If a sequence is Cauchy, then it is bounded.
If $X_n$ is a sequence of real numbers such that $|X_n| \leq K$ for all $n$, then $X_n$ is a bounded sequence.
A sequence $X$ is bounded if and only if there exists a constant $K > 0$ such that $|X_n| \leq K$ for all $n$.
If $X$ is a bounded sequence, then for every $K > 0$, if $\|X_n\| \leq K$ for all $n$, then $Q$.
If $X$ is a bounded sequence, then there exists a positive real number $K$ such that $|X_n| \leq K$ for all $n$.
If $X_n$ is a sequence of complex numbers such that $|X_n| \leq K$ for all $n$, then $X_n$ is a bounded sequence.
If $X$ is a bounded sequence, then the range of $X$ is bounded above.
If $X$ is a bounded sequence, then the sequence of norms of $X$ is bounded above.
If $X$ is a bounded sequence, then the range of $X$ is bounded below.
If $f_n$ is eventually bounded above by $g_n$ and $g_n$ is bounded, then $f_n$ is bounded.
A sequence of vectors is bounded if and only if there exists a natural number $N$ such that the norm of each vector in the sequence is less than or equal to $N + 1$.
A sequence $X$ is bounded if and only if there exists a natural number $N$ such that $|X_n| \leq N$ for all $n$.
A sequence of vectors is bounded if and only if there exists a natural number $N$ such that the norm of each vector in the sequence is less than $N + 1$.
A sequence $X$ is bounded if and only if there exists a natural number $N$ such that $|X_n| < N + 1$ for all $n$.
A sequence $X$ is bounded if and only if there exists a constant $k > 0$ and a point $x$ such that $|X_n - x| \leq k$ for all $n$.
A sequence $X$ is bounded if and only if there exists a constant $k > 0$ and a number $N$ such that for all $n$, we have $|X_n - X_N| \leq k$.
A sequence of vectors is bounded if and only if its negation is bounded.
If $f$ is a bounded sequence, then so is $f + c$.
A sequence $f$ is bounded if and only if the sequence $f + c$ is bounded for any constant $c$.
If $f$ and $g$ are bounded sequences, then $f \cdot g$ is a bounded sequence.
The function that maps every element of a set $F$ to a constant $c$ is a Baire function.
If $c \neq 0$, then a sequence $f$ is bounded if and only if the sequence $c f$ is bounded.
If $f$ is a bounded sequence, then so is $f \circ g$ for any function $g$.
A sequence $f$ is bounded if and only if the sequence $f(n+1)$ is bounded.
If $f$ is a bounded sequence and $g$ is a strictly increasing sequence of natural numbers, then the sequence $f \circ g$ is bounded if and only if $f$ is bounded.
If $f$ is a nonnegative increasing sequence, then $f$ is a bounded sequence if and only if the subsequence $f \circ g$ is a bounded sequence.
If the range of a sequence is contained in a bounded interval, then the sequence is bounded.
If $X$ is an increasing sequence of real numbers and $X_i \leq B$ for all $i$, then $X$ is bounded.
If $X$ is a decreasing sequence of real numbers and $X_i \geq B$ for all $i$, then $X$ is bounded.
If $f$ is a polynomial function, then there exists a constant $M$ such that for all $z$ with $|z| > M$, we have $|f(z)| \leq e |z|^{n+1}$.
If $c_k \neq 0$ and $1 \leq k \leq n$, then the function $f(z) = \sum_{i=1}^n c_i z^i$ is unbounded on the complex plane.
If $f$ is a function such that for every $r > 0$, there exists an $x$ such that $|f(x)| < r$, then $f$ is a zero function.