Two Different Notions of Smallness
In a topological space, meagerness measures how a set sits among nowhere-dense sets. Lebesgue measure instead measures the total length needed to cover a set. These notions often describe sets that seem small, but they do not imply one another. We will compare them in the unit interval \(I=[0,1]\), with its usual subspace topology from \(\mathbb{R}\) and Lebesgue measure.
A subset \(A\subseteq I\) is nowhere dense in \(I\) if its closure in \(I\) has empty interior in \(I\). It is meager in \(I\) if it is a countable union of nowhere-dense subsets of \(I\). A set \(A\subseteq I\) has Lebesgue measure zero if, for every \(\varepsilon>0\), it can be covered by countably many intervals in \(\mathbb{R}\) whose total length is less than \(\varepsilon\). We write \(m(A)\) for its Lebesgue measure when that measure is being used.
Meagerness is unchanged if a nowhere-dense set is replaced by its closure, so a meager set is contained in a countable union of closed nowhere-dense sets. For measure, we use the basic properties of Lebesgue measure: countable subadditivity, countable additivity for disjoint measurable sets, and the fact that an interval has measure equal to its length. The category and measure definitions involve different tests: one concerns interiors of closures, the other concerns lengths of covers.
The Rationals: Small in Both Senses
Worked Example: The Rationals in the Unit Interval
Let \(A=\mathbb{Q}\cap I\), where \(\mathbb{Q}\) is the set of rational numbers. We check separately that \(A\) is meager and has measure zero.
Meagerness. The set \(A\) is countable, so write \(A=\bigcup_{q\in A}\{q\}\). Each singleton \(\{q\}\) is closed in \(I\). It has empty interior: every relatively open neighborhood of any point \(q\in I\) contains points of \(I\) other than \(q\). Thus each singleton is nowhere dense, and \(A\) is meager.
Measure zero. Enumerate \(A\) as \((q_j)_{j\in\mathbb{N}}\). Given \(\varepsilon>0\), cover \(q_j\) by the interval centered at \(q_j\) with radius \(r_j=\varepsilon/2^{j+2}\). Its length is \(2r_j=\varepsilon/2^{j+1}\), so the total length of these intervals is \(\sum_{j=1}^{\infty}\varepsilon/2^{j+1}=\varepsilon/2<\varepsilon\). This works for every positive \(\varepsilon\), so \(A\) has measure zero.
Check. The radii are positive, each \(q_j\) lies in its covering interval, and the geometric series used for the total length is \(\sum_{j=1}^{\infty}2^{-(j+1)}=1/2\). Thus both claims follow from the respective definitions.
This example confirms that a set can be small in both senses. It does not show that either notion implies the other: to test an implication, we need examples where one property holds and the other fails.
Meager but Not Measure Zero
Worked Example: A Nowhere-Dense Set of Positive Measure
We construct a closed set \(K\subseteq I\) by repeatedly deleting open middle intervals. At stage \(n\in\mathbb{N}\), remove from each of the \(2^{n-1}\) remaining closed intervals its open middle interval of length \(4^{-n}\). Let \(F_n\) be the set remaining after stage \(n\), and put \(F_0=I\) and \(K=\bigcap_{n=0}^{\infty}F_n\).
We first verify that the construction is possible and compute the total length remaining at each stage. At stage \(n\), the total length removed is \(2^{n-1}4^{-n}=2^{-(n+1)}\). Consequently, the total length of \(F_n\) is \(1-\sum_{k=1}^{n}2^{-(k+1)}=1/2+2^{-(n+1)}\). There are \(2^n\) intervals in \(F_n\), all of the same length, so each has length \((1/2+2^{-(n+1)})/2^n=2^{-(n+1)}+2^{-2n-1}\). In particular, before stage \(n\), each interval has length \(2^{-n}+2^{-2n+1}\), which is greater than the length \(4^{-n}=2^{-2n}\) to be removed. The deleted middle interval therefore fits inside each interval at every stage.
Each \(F_n\) is a finite union of closed intervals, hence closed in \(I\), so \(K\) is closed in \(I\). The deleted intervals are pairwise disjoint, and their total length is \(\sum_{n=1}^{\infty}2^{-(n+1)}=1/2\). Their union is \(I\setminus K\). Countable additivity, together with \(m(I)=1\), gives \(m(K)=1-1/2=1/2\). Thus \(K\) does not have measure zero.
To check that \(K\) is nowhere dense, note that each component interval of \(F_n\) has length \(2^{-(n+1)}+2^{-2n-1}\), which tends to zero as \(n\) increases. If \(K\) contained a nonempty relatively open interval of \(I\), that interval would have some positive length \(\delta\). Choose \(n\) so large that every component of \(F_n\) has length less than \(\delta\). Since \(K\subseteq F_n\), the interval would have to lie in one component of \(F_n\): an interval cannot cross the gap between two distinct components while remaining in \(F_n\). This is impossible because its length is greater than the length of that component. Hence \(K\) has empty interior in \(I\); as \(K\) is closed, it is nowhere dense and therefore meager.
Check. We have proved directly that \(K\) is closed with empty interior, and computed \(m(K)=1/2>0\). So meagerness does not imply measure zero.
Measure Zero but Not Meager
For the reverse failure, we construct a dense \(G_\delta\) set of measure zero. A \(G_\delta\) set is a countable intersection of open sets. We use the rationals as a countable dense set in \(I\), but make the neighborhoods covering them sufficiently short that their total length is small.
Worked Example: A Dense \(G_\delta\) Set of Measure Zero
Enumerate \(\mathbb{Q}\cap I\) without repetitions as \((q_j)_{j\in\mathbb{N}}\). For \(n,j\in\mathbb{N}\), set \(r_{n,j}=2^{-(n+j+2)}\), and define the relatively open set \(U_n=I\cap\bigcup_{j=1}^{\infty}(q_j-r_{n,j},q_j+r_{n,j})\). Every rational in \(I\) belongs to \(U_n\), so \(U_n\) is dense in \(I\): every nonempty relatively open subset of \(I\) meets \(\mathbb{Q}\cap I\). It is open in \(I\) because it is the intersection of \(I\) with an open subset of \(\mathbb{R}\).
The intervals used to cover \(U_n\) have total length \(\sum_{j=1}^{\infty}2r_{n,j}=\sum_{j=1}^{\infty}2^{-(n+j+1)}=2^{-(n+1)}\). Now let \(G=\bigcap_{n=1}^{\infty}U_n\). This is a \(G_\delta\) set. Since \(G\subseteq U_n\) for every \(n\), for each \(\varepsilon>0\) we can choose \(n\) with \(2^{-(n+1)}<\varepsilon\), and use the displayed cover of \(U_n\) to cover \(G\) by intervals of total length less than \(\varepsilon\). Therefore \(G\) has measure zero.
It remains to prove that \(G\) is not meager. We use the Baire category theorem: a complete metric space is Baire, meaning every countable intersection of dense open sets is dense. This theorem was established in the earlier material on the Baire category theorem. The interval \(I\) with its usual metric is complete, so \(G\), the intersection of the dense open sets \(U_n\), is dense in \(I\). Also \(I\setminus G=\bigcup_{n=1}^{\infty}(I\setminus U_n)\), and each \(I\setminus U_n\) is closed with empty interior because \(U_n\) is open and dense. Thus \(I\setminus G\) is meager.
If \(G\) were meager as well, then \(I=G\cup(I\setminus G)\) would be meager. That contradicts the Baire theorem: \(I\) is nonempty and is not meager in itself. Hence \(G\) is not meager. We have constructed a measure-zero set that is not meager.
Check. The same explicit radii give a cover of \(G\) of total length at most \(2^{-(n+1)}\) for every \(n\), so the bound tends to zero. The nonmeagerness conclusion uses completeness through the Baire theorem, not density alone.
What the Examples Establish
Proof. The set \(K\) constructed above is nowhere dense, hence meager, and satisfies \(m(K)=1/2\), so it disproves the first implication. The set \(G\) constructed above has measure zero and is not meager, so it disproves the second. Both examples are subsets of the same space \(I\), with the topology and measure specified here. \(\square\)
The distinction is useful whenever a result describes a “typical” point or a “small” exceptional set. A meager set may have positive measure, and a set of measure zero may be topologically large enough to be nonmeager. One should therefore not replace a category conclusion with a measure conclusion, or conversely, without a theorem that supplies the required connection.
Common pitfall. From the fact that \(\mathbb{Q}\cap I\) is both meager and measure zero, one might conclude that these notions of smallness coincide. That example shows only that the two properties can occur together. The set \(K\) shows that meager sets need not have measure zero; \(G\) shows that measure-zero sets need not be meager.
Check Your Understanding
Use the definitions and the arguments above to answer the following questions.
- Why does countability imply that \(\mathbb{Q}\cap I\) is meager in \(I\)? What property of the relative topology on \(I\) is needed for the singleton sets to be nowhere dense?
- At stage \(n\) of the construction of \(K\), how many intervals are removed, and what is their total length?
- Why does the fact that the component lengths of \(F_n\) tend to zero imply that \(K\) has empty interior?
- In the construction of \(G\), verify that \(\sum_{j=1}^{\infty}2^{-(n+j+1)}=2^{-(n+1)}\).
- Which Baire-space fact is needed to conclude that the measure-zero set \(G\) is not meager?