Measure Bounds for Rational Translates

From Rational Translates to Measure Bounds

Let $V\subseteq[0,1]$ be a set containing exactly one representative from each equivalence class in $[0,1]$, where two points are equivalent when their difference is rational. The rational translates of $V$ form a countable family with two useful geometric properties: they are pairwise disjoint, and a suitable subfamily covers $[0,1]$. If we additionally assume that $V$ is Lebesgue measurable, translation invariance gives the same measure to every translate. Countable additivity can then convert the geometric properties into bounds on a sum of equal terms.

Throughout, $m$ denotes Lebesgue measure on $\mathbb{R}$, and $\mathcal{L}$ denotes the Lebesgue sigma-algebra. The set $V$ is assumed to be in $\mathcal{L}$ for the measure comparisons below. In particular, $0\le m(V)\le 1$, since $V\subseteq[0,1]$. Write $\mathbb{Q}$ for the rational numbers and set $D=\mathbb{Q}\cap[-1,1]$. For each $q\in D$, define $V+q=\{v+q:v\in V\}$.

Key setup: If $V$ contains exactly one representative from each rational-equivalence class in $[0,1]$, then the sets $V+q$, for $q\in D$, are pairwise disjoint, their union contains $[0,1]$, and their union is contained in $[-1,2]$. The intervals have measures $m([0,1])=1$ and $m([-1,2])=3$.

Here are the geometric facts in the form needed for the measure argument. To see coverage, fix $x\in[0,1]$, and let $v\in V$ be the representative of the equivalence class of $x$. Then $x-v\in\mathbb{Q}$, and $x,v\in[0,1]$ implies $-1\le x-v\le1$. Thus $q=x-v\in D$ and $x=v+q\in V+q$. To see disjointness, suppose $v+q=w+r$ for $v,w\in V$ and $q,r\in D$. Then $v-w=r-q\in\mathbb{Q}$, so $v$ and $w$ belong to the same equivalence class. The representative property gives $v=w$, and the original equality then gives $q=r$. Finally, $0\le v\le1$ and $-1\le q\le1$ imply $-1\le v+q\le2$.

Translation Invariance and Measurability

The measure argument needs more than geometric disjointness: each set in the family must be measurable and have the same measure. We use translation invariance of Lebesgue measure. This property follows directly from its construction. Lebesgue outer measure $m^*$ is defined using countable interval covers and the sums of their lengths. Translating an interval cover by $t\in\mathbb{R}$ preserves every interval length, and translating back gives the reverse comparison. Hence $m^*(A+t)=m^*(A)$ for every $A\subseteq\mathbb{R}$. The Carathéodory criterion for Lebesgue measurability is preserved under translation: applying the criterion to $A$ with the test set $S-t$ gives the criterion for $A+t$ with test set $S$. Therefore, if $A\in\mathcal{L}$, then $A+t\in\mathcal{L}$ and $m(A+t)=m(A)$.

Lemma (Translation invariance): If $A\in\mathcal{L}$ and $t\in\mathbb{R}$, then $A+t\in\mathcal{L}$ and $m(A+t)=m(A)$.

Proof. As described above, translating any interval cover of $A$ gives an interval cover of $A+t$ with exactly the same sum of lengths. Taking the infimum over covers gives $m^*(A+t)\le m^*(A)$. Translating covers of $A+t$ by $-t$ gives $m^*(A)\le m^*(A+t)$, so the outer measures are equal. If $A$ is Lebesgue measurable, the Carathéodory criterion says that for every $T\subseteq\mathbb{R}$, $m^*(T)=m^*(T\cap A)+m^*(T\setminus A)$. Given $S\subseteq\mathbb{R}$, apply this equality with $T=S-t$. Translation preserves outer measure and maps the two sets on the right to $S\cap(A+t)$ and $S\setminus(A+t)$, respectively. Thus $A+t$ satisfies the criterion and is measurable. Its measure, which equals its outer measure, is $m^*(A+t)=m^*(A)=m(A)$. $\square$

Since $V\in\mathcal{L}$, the lemma shows that $V+q\in\mathcal{L}$ and $m(V+q)=m(V)$ for every $q\in D$. The index set $D$ is countable, so the union $U=\bigcup_{q\in D}(V+q)$ is measurable. The disjointness proved above allows countable additivity to be applied to this union.

The Measure Bounds

Proposition (Bounds for rational translates): Under the assumptions above, let $c=m(V)$ and $U=\bigcup_{q\in D}(V+q)$. Then $$ 1\ \le\ m(U)\ =\ \sum_{q\in D}m(V+q)\ =\ \sum_{q\in D}c\ \le\ 3. $$ The sum is a countable sum of nonnegative terms. The inequalities use $[0,1]\subseteq U\subseteq[-1,2]$.

Proof. Each $V+q$ is measurable by translation invariance, and the sets are pairwise disjoint. Countable additivity therefore gives $m(U)=\sum_{q\in D}m(V+q)$. Translation invariance gives $m(V+q)=c$ for every $q\in D$, so this is $\sum_{q\in D}c$. Because $[0,1]\subseteq U$, monotonicity of Lebesgue measure gives $m([0,1])\le m(U)$. The interval has measure $1-0=1$, so $1\le m(U)$. Because $U\subseteq[-1,2]$, monotonicity also gives $m(U)\le m([-1,2])=2-(-1)=3$. Combining these equalities and inequalities proves the assertion. $\square$

These estimates are bounds on the measure of the full countable union. In particular, the lower bound comes from coverage by all the translates indexed by $D$; it is not a claim that an arbitrary finite selection of translates covers $[0,1]$. The upper bound comes from one common containing interval for the full family.

Worked Example: Four Disjoint Translates and the Common Upper Bound

Take the four distinct rational numbers $q_1=-1$, $q_2=-\tfrac13$, $q_3=\tfrac13$, and $q_4=1$, all of which belong to $D$. Under the assumption that $V$ is measurable, the four sets $V+q_j$ are measurable and pairwise disjoint. If $c=m(V)$, translation invariance gives measure $c$ to each one. Finite additivity, which follows from countable additivity, gives

m\left(\bigcup_{j=1}^{4}(V+q_j)\right)=c+c+c+c=4c.

Every one of these translates lies in $[-1,2]$, so their union lies there as well. Monotonicity and the interval length give $4c\le m([-1,2])=2-(-1)=3$, hence $c\le\tfrac34$. Check: if $c=\tfrac12$, the union would have measure $4\cdot\tfrac12=2$, which satisfies $2\le3$. If $c=1$, the union would have measure $4\cdot1=4$, which violates the required bound $4\le3$. This check concerns this four-set subfamily only; it does not give it coverage of $[0,1]$.

Worked Example: Testing a Proposed Positive Measure Against Finite Subfamilies

Suppose, for a bound check, that the common measure were $c=\tfrac1{10}$. Choose $31$ distinct rationals $q_j=-1+\tfrac{j}{15}$ for $j=0,1,\ldots,30$. The first is $-1$, the last is $-1+\tfrac{30}{15}=1$, and successive values differ by $\tfrac1{15}>0$; thus all $31$ are distinct and lie in $D$. Their translates are pairwise disjoint. The measure of their union would be

31c=31\cdot\frac1{10}=\frac{31}{10}.

But the union is contained in $[-1,2]$, whose measure is $2-(-1)=3=\tfrac{30}{10}$. The proposed value would require $\tfrac{31}{10}\le\tfrac{30}{10}$, which is false. Thus $c=\tfrac1{10}$ cannot satisfy the upper bound for this family. Check: the comparison is exact: $\tfrac{31}{10}-\tfrac{30}{10}=\tfrac1{10}>0$. This argument uses only containment and finite additivity; it does not use the coverage lower bound.

Worked Example: Why the Full Union Supplies the Lower Bound

Coverage gives $[0,1]\subseteq U$, so monotonicity yields $m(U)\ge m([0,1])=1-0=1$. The full union is contained in $[-1,2]$, giving $m(U)\le m([-1,2])=2-(-1)=3$. Countable additivity identifies the same quantity as $\sum_{q\in D}c$. For comparison, the single translate with $q=0$ is just $V\subseteq[0,1]$, and its measure satisfies only $0\le c\le1$; that one set is not asserted to cover $[0,1]$.

For an explicit finite calculation, use $q=-1,0,1$. The three translates are disjoint and have total measure $c+c+c=3c$, with $3c\le3$, so $c\le1$. This finite calculation gives no lower bound of $1$ for $3c$: the three translates need not cover $[0,1]$. The lower bound $1\le m(U)$ is justified by coverage by the entire family indexed by $D$, not by these three selected terms. Check: the containing-interval estimate for this finite union is $3c\le m([-1,2])=3$, and the full-union coverage estimate is $m(U)\ge m([0,1])=1$; the two estimates apply to different set inclusions.

What the Bounds Do—and Do Not—Say

The key output is the pair of inequalities for one and the same measurable union: $1\le m(U)\le3$. Countable additivity also expresses $m(U)$ as a sum with one identical term for each rational $q\in D$. The countability of $D$ matters because countable additivity applies to countable disjoint families; the uncountability of $V$ is not an obstacle to using the measure of each translate.

A common mistake is to infer that every finite subfamily covers $[0,1]$ just because the full family does. That would incorrectly give a lower bound for the measure of a finite union. For example, $V+0=V$ is contained in $[0,1]$, but the representative property alone does not say $V=[0,1]$. Another error is to use the length $2$ for the containing interval: the correct interval is $[-1,2]$, whose length is $2-(-1)=3$, not $2$. The endpoints and the full interval must be included when applying monotonicity.

The bounds are obtained under the explicit assumption $V\in\mathcal{L}$. They do not assert measurability of an arbitrary subset of $[0,1]$, and the measure equalities are not statements merely holding almost everywhere: they are exact equalities of measures of measurable sets. The geometric coverage and disjointness identify the relevant union, while translation invariance and countable additivity supply the measure calculation.

Check Your Understanding

For each question, distinguish the full countable union from a finite subfamily and state which measure property justifies the step.

Why does $x\in[0,1]$ belong to $V+q$ for some $q\in D$, and why is that $q$ in $[-1,1]$?
Given $v+q=w+r$ with $v,w\in V$ and $q,r\in D$, identify the rational difference that forces $v=w$.
If $m(V)=c$, what is the measure of the union of $N$ distinct translates, and which two facts justify the answer?
Explain why the inclusion $[0,1]\subseteq U$ gives a lower bound for $m(U)$, but does not automatically give the same lower bound for an arbitrary finite subfamily.
Compute the length of $[-1,2]$ and state precisely how its containment of $U$ yields the upper bound.

Measure Bounds for Rational Translates

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