From Distance Control to Hausdorff Measure
A Lipschitz map controls how much the diameter of a set can grow. Hausdorff measure is built from covers and their diameters, so this distance control gives a direct comparison between the measure of a set and that of its image. The same comparison then constrains Hausdorff dimension. These statements apply to maps between general metric spaces; no linear structure or differentiability is needed.
Let \((X,d_X)\) be a metric space and \(A\subseteq X\). For \(s>0\) and \(\delta>0\), define the \(s\)-dimensional Hausdorff content at scale \(\delta\) by
Here the covers are countable families of subsets of \(X\), and \(\operatorname{diam} U=\sup\{d_X(x,y):x,y\in U\}\). Empty members can be discarded. The Hausdorff outer measure is \(\mathcal{H}^s(A)=\lim_{\delta\downarrow0}\mathcal{H}^s_\delta(A)\); the limit exists in \([0,\infty]\) because the contents are nondecreasing as \(\delta\) decreases. We use the convention that a nonempty set contributes \(1\) to a cover when \(s=0\), so \(\mathcal{H}^0\) counts points. Hausdorff dimension is
with \(\dim_H\varnothing=0\). Hausdorff measure here is initially an outer measure, so these definitions make sense for every subset, without assuming measurability.
The Lipschitz Measure Inequality
Proof. Fix \(s>0\), \(\delta>0\), and a countable cover \((U_j)\) of \(E\) by subsets of \(X\) with \(\operatorname{diam}U_j\le\delta\). Replace each member by \(U_j\cap E\), and discard empty intersections. The sets \(f(U_j\cap E)\) cover \(f(E)\). For any two points \(f(x),f(y)\) in one such image, the Lipschitz condition gives \(d_Y(f(x),f(y))\le Ld_X(x,y)\). Taking the supremum over pairs shows \(\operatorname{diam}f(U_j\cap E)\le L\operatorname{diam}(U_j\cap E)\le L\delta\). Moreover, \[ \sum_j\bigl(\operatorname{diam}f(U_j\cap E)\bigr)^s \le L^s\sum_j(\operatorname{diam}U_j)^s. \] Taking the infimum over the original covers yields \(\mathcal{H}^s_{L\delta}(f(E))\le L^s\mathcal{H}^s_\delta(E)\). As \(\delta\downarrow0\), also \(L\delta\downarrow0\); taking limits gives the asserted inequality, including when either side is infinite. For \(s=0\), each nonempty \(U_j\cap E\) has a nonempty image, and the same cover argument does not increase the number of covering sets. Thus \(\mathcal{H}^0(f(E))\le\mathcal{H}^0(E)\), which is the stated bound because \(L^0=1\). \(\square\)
The theorem also has a zero-Lipschitz case, which is best stated separately rather than by inserting \(L=0\) into the formula. If \(E\ne\varnothing\) and \(d_Y(f(x),f(y))\le0\) for all \(x,y\), then all image points coincide, so \(f(E)\) is a singleton. Consequently \(\mathcal{H}^s(f(E))=0\) for \(s>0\), and \(\dim_H f(E)=0\). If \(E=\varnothing\), its image is empty. For \(s=0\), a nonempty domain and its singleton image both have \(\mathcal{H}^0=1\).
Worked Example: A Similarity Scales Hausdorff Measure
Consider \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=3x+2\), with the usual distance. For any \(x,y\in\mathbb{R}\), \(|f(x)-f(y)|=|3x+2-(3y+2)|=3|x-y|\). Thus \(f\) is Lipschitz with \(L=3\), and the theorem gives \(\mathcal{H}^s(f(E))\le3^s\mathcal{H}^s(E)\).
There is also reverse distance control: \(|x-y|=\tfrac13|f(x)-f(y)|\). The inverse on \(f(E)\) is therefore Lipschitz with constant \(1/3\). Applying the theorem to that inverse gives \(\mathcal{H}^s(E)\le(1/3)^s\mathcal{H}^s(f(E))\). Combining both inequalities, \(\mathcal{H}^s(f(E))=3^s\mathcal{H}^s(E)\) for \(s>0\), with equality interpreted in the extended nonnegative reals. For a concrete check, take \(E=\{0,1\}\). Then \(f(E)=\{2,5\}\); both sets have \(\mathcal{H}^0=2\), and both have \(\mathcal{H}^s=0\) for every \(s>0\). The formula gives \(3^s\cdot0=0\) for \(s>0\), as required.
Worked Example: Checking the Cover Estimate for a Contraction
Let \(E=[0,1]\subset\mathbb{R}\), \(f(x)=x/4+1\), and \(s=2\). The distance calculation \(|f(x)-f(y)|=|x-y|/4\) shows that \(L=1/4\). Divide \(E\) into four closed intervals \(I_k=[k/4,(k+1)/4]\), for \(k=0,1,2,3\). They cover \(E\), and each has diameter \(1/4\). Their images are four closed intervals of diameter \(1/16\), which cover \(f(E)=[1,5/4]\). The source cover sum is \(4(1/4)^2=4/16=1/4\). The image cover sum is \(4(1/16)^2=4/256=1/64\). The ratio is \((1/64)/(1/4)=1/16=(1/4)^2=L^2\), exactly the factor predicted by the theorem. This is a comparison of cover sums at these scales, not a claim that either particular cover sum equals the corresponding Hausdorff measure. The check also confirms the direction of the factor: contraction by \(1/4\) multiplies squared diameters by \(1/16\).
Hausdorff Dimension Cannot Increase
Proof. If \(E=\varnothing\), both dimensions are zero. If \(L=0\) and \(E\ne\varnothing\), the image is a singleton and has dimension zero, so the assertion holds. Now suppose \(L>0\). The standard critical-exponent property of Hausdorff measure says that if \(t>\dim_H E\), then \(\mathcal{H}^t(E)=0\); this is the zero-measure side of the critical-exponent characterization established in the earlier material on Hausdorff dimension. The measure inequality then gives \(\mathcal{H}^t(f(E))\le L^t\mathcal{H}^t(E)=0\). Hence \(\dim_H f(E)\le t\) for every \(t>\dim_H E\), and letting \(t\) decrease to \(\dim_H E\) proves the claim. If \(\dim_H E=\infty\), the inequality is automatic in the extended order. \(\square\)
In particular, if \(L\le1\), the map does not increase Hausdorff measure at any positive dimension, and it never increases Hausdorff dimension. For \(L>1\), the measure may increase by the factor \(L^s\); the dimension conclusion still holds because multiplication by a finite positive constant does not change which Hausdorff measures are zero.
Worked Example: Projection Does Not Increase Dimension
For \(E\subseteq\mathbb{R}^2\), let \(\pi:E\to\mathbb{R}\) be projection onto the first coordinate, \(\pi(x_1,x_2)=x_1\). For points \(x=(x_1,x_2)\) and \(y=(y_1,y_2)\), \(|\pi(x)-\pi(y)|=|x_1-y_1|\le\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=|x-y|\). Thus \(\pi\) is \(1\)-Lipschitz and \(\mathcal{H}^s(\pi(E))\le\mathcal{H}^s(E)\) for every \(s>0\), so \(\dim_H\pi(E)\le\dim_HE\).
For a numerical distance check, take \(x=(1,2)\), \(y=(4,6)\). Their distance is \(\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=5\), while the projected distance is \(|4-1|=3\), and \(3\le5\). Projection can identify points: \((1,2)\) and \((1,7)\) both map to \(1\). This does not invalidate the cover argument; image sets can have smaller diameter or coincide, which only makes the image cover no more costly.
Bi-Lipschitz Maps Preserve Dimension
Proof. The upper Lipschitz bound and the theorem give the right-hand measure inequality. Apply the theorem to the inverse, which has Lipschitz constant \(1/c\): \(\mathcal{H}^s(E)\le c^{-s}\mathcal{H}^s(f(E))\). Multiplying by \(c^s>0\) gives the left-hand inequality. The dimension corollary applied to \(f\) gives \(\dim_H f(E)\le\dim_H E\); applied to \(f^{-1}\), it gives the reverse inequality. Therefore the dimensions are equal. \(\square\)
Two-sided distance control is a sufficient way to prove equality, not a necessary condition for equality. A map may fail to have a Lipschitz inverse while its image nevertheless has the same dimension as its domain.
Worked Example: Equal Dimensions Without a Lipschitz Inverse
Consider \(f:[0,1]\to[0,1]\), \(f(x)=x^2\). It is Lipschitz with constant \(2\), since \(|x^2-y^2|=|x-y||x+y|\le2|x-y|\) for \(x,y\in[0,1]\). It is onto \([0,1]\): for each \(u\in[0,1]\), \(x=\sqrt{u}\in[0,1]\) satisfies \(f(x)=u\). Thus \(f([0,1])=[0,1]\), so the domain and image are the same set and have equal Hausdorff dimensions.
However, the inverse \(g(u)=\sqrt{u}\) is not Lipschitz. If it had a finite Lipschitz constant \(K\), comparing \(u=1/n^2\) and \(v=0\) for each positive integer \(n\) would give \(|g(1/n^2)-g(0)|=1/n\le K/n^2\), hence \(n\le K\) for every positive integer \(n\), which is impossible. This example verifies why reverse distance control cannot be asserted as necessary for equality of dimensions.
A Common Pitfall
A frequent error is to reverse the factor in the measure inequality. If a map is \(L\)-Lipschitz, a cover member of diameter \(r\) has image diameter at most \(Lr\), so its contribution to an \(s\)-dimensional cover sum is at most \(L^sr^s\), not \(L^{-s}r^s\). The contraction example makes the distinction exact: its factor is \((1/4)^2=1/16\), not \(16\).
A second error is to infer that a Lipschitz map must strictly lower dimension, or that equality requires a Lipschitz inverse. Neither follows. An injective Lipschitz map can preserve dimension, and the map \(x\mapsto x^2\) above preserves the dimension of its domain despite its non-Lipschitz inverse. The conclusions to retain are precise: Lipschitz maps cannot increase Hausdorff dimension, while bi-Lipschitz maps preserve it. Reverse distance control supplies a proof of equality, but is not a necessary condition.
Check Your Understanding
For each question, identify the metric spaces and verify the distance or covering condition before applying a conclusion.
- If \(f\) is \(L\)-Lipschitz with \(L>0\), explain why the image of a cover at scale \(\delta\) is a cover at scale \(L\delta\).
- For \(f(x)=5x-1\) on \(\mathbb{R}\), what is the exact factor relating \(\mathcal{H}^s(f(E))\) and \(\mathcal{H}^s(E)\) when \(s>0\)?
- Why does a constant map from a nonempty set have image of Hausdorff dimension zero, even if its domain has infinite dimension?
- State the two inequalities needed to show that a bi-Lipschitz map preserves Hausdorff dimension.
- Does equality of the dimensions of a set and its image imply that the inverse map is Lipschitz? Use \(x\mapsto x^2\) on \([0,1]\) to justify your answer.