Hausdorff Dimension of the Brownian Zero Set

The Zero Set and Hausdorff Dimension

Brownian motion has zeros arbitrarily close to its starting time, but that fact alone does not describe the size of the set of zero times. Hausdorff dimension provides a way to measure that size by asking how efficiently a set can be covered by small intervals. For standard Brownian motion, the zero set on a bounded time interval has dimension one half almost surely. We will prove the upper bound by counting dyadic intervals that meet the zero set, and the lower bound by using local time to construct a measure with finite energy.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) carrying a standard Brownian motion \((B_t)_{t\ge0}\), with \(B_0=0\). We focus on the random subset \(Z=\{t\in[0,1]:B_t=0\}\) of the time axis. Brownian paths are continuous, so \(Z\) is closed and hence compact.

Definition: For a set \(A\subseteq\mathbb{R}\), a number \(s\ge0\), and \(\delta>0\), let \(\mathcal{H}^s_\delta(A)\) be the infimum of \(\sum_i\operatorname{diam}(U_i)^s\) over countable covers \(A\subseteq\bigcup_i U_i\) with \(\operatorname{diam}(U_i)\le\delta\). The \(s\)-dimensional Hausdorff measure is \(\mathcal{H}^s(A)=\lim_{\delta\downarrow0}\mathcal{H}^s_\delta(A)\), and the Hausdorff dimension is \(\dim_H(A)=\inf\{s:\mathcal{H}^s(A)=0\}\).

We use the convention that a set of diameter zero contributes zero when \(s>0\). A useful way to prove a dimension lower bound is the energy method: find a nonzero finite measure supported on the set whose \(s\)-energy is finite. The measure need not be defined on every arbitrary cover set; in the proof below, outer measure will handle that issue.

The Upper Bound from Dyadic Covers

For \(n\ge1\), set \(\delta_n=2^{-n}\), and let \(N_n\) be the number of intervals \([k\delta_n,(k+1)\delta_n]\), \(0\le k<2^n\), that intersect \(Z\). These intervals cover \(Z\), and all have diameter \(\delta_n\). We first bound the expected number of intervals needed.

Proposition: For every \(n\ge1\), \(\mathbb{E}N_n\le 1+\frac{4}{\pi}2^{n/2}\). Consequently, \(\dim_H(Z)\le\tfrac12\) almost surely.

Proof. Write \(\delta=2^{-n}\) and \(m=2^n\). The interval with \(k=0\) meets \(Z\), since \(B_0=0\), so it contributes exactly one. For \(1\le k<m\), condition on \(B_{k\delta}=x\). The increment over the next interval is independent of the past and is Brownian motion over length \(\delta\). The reflection principle gives probability \(2\Phi(-|x|/\sqrt{\delta})\) of hitting zero during that interval, where \(\Phi\) is the standard normal distribution function. Since \(B_{k\delta}/\sqrt{\delta}\) has density \(y\mapsto (2\pi k)^{-1/2}e^{-y^2/(2k)}\), whose maximum is \((2\pi k)^{-1/2}\),

\mathbb{P}([k\delta,(k+1)\delta]\cap Z\ne\varnothing) \le \frac{1}{\sqrt{2\pi k}}\int_{\mathbb{R}}2\Phi(-|y|)\,dy =\frac{2}{\pi\sqrt{k}}.

For the integral, \(\int_0^\infty\Phi(-y)\,dy=1/\sqrt{2\pi}\), obtained by integrating the standard normal density over the region \(0\le y\le z\) and reversing the order of integration. Thus the integral over \(\mathbb{R}\) above is \(4/\sqrt{2\pi}\), giving the displayed bound. Also, \(\sum_{k=1}^{m-1} k^{-1/2}\le 2\sqrt m\): for each \(k\ge1\), \(k^{-1/2}\le\int_{k-1}^{k}x^{-1/2}\,dx\), with the integral at zero understood as an improper integral. Therefore \(\mathbb{E}N_n\le1+(4/\pi)\sqrt m=1+(4/\pi)2^{n/2}\).

Now fix \(s>\tfrac12\). The covering definition gives \(\mathcal{H}^s_{\delta_n}(Z)\le N_n\delta_n^s\). Moreover,

\mathbb{E}\!\left[\sum_{n=1}^{\infty}N_n2^{-ns}\right] \le\sum_{n=1}^{\infty}\left(1+\frac{4}{\pi}2^{n/2}\right)2^{-ns}<\infty,

because both geometric series converge when \(s>\tfrac12\). A nonnegative random variable with finite expectation is finite almost surely, so \(N_n2^{-ns}\to0\) almost surely. Since \(\delta_n\downarrow0\), this implies \(\mathcal{H}^s(Z)=0\) almost surely. Applying this conclusion to the countable sequence \(s=\tfrac12+1/j\), \(j\ge1\), gives \(\dim_H(Z)\le\tfrac12\) almost surely. \(\square\)

Worked Example: Counting at a Fixed Dyadic Scale

At \(n=2\), the mesh is \(\delta=1/4\), so there are four intervals. The first meets \(Z\) at time zero. For the other three, the respective bounds above, using \(k=1,2,3\), are \(2/\pi\), \(2/(\pi\sqrt2)\), and \(2/(\pi\sqrt3)\). Thus

\mathbb{E}N_2 \le 1+\frac{2}{\pi}\left(1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\right).

This is an upper bound, not an exact expected count: the events for different intervals need not be independent, and the separate probability bounds need not be equalities. The cover at this scale has total \(s\)-cost at most \(N_2(1/4)^s\), which is the quantity used across finer scales in the proposition.

The Lower Bound from Local Time

For the lower bound, the relevant measure lives on the time axis and is generated by Brownian local time at zero. We state the local-time input precisely. This is a standard Brownian local-time theorem; its proof is beyond this tutorial.

Standard Brownian local-time result: There is a continuous increasing local time \(L_t^x\), with the occupation-density normalization \(\int_0^t f(B_u)\,du=\int_{\mathbb{R}}f(x)L_t^x\,dx\) for bounded nonnegative Borel \(f\). For \(x=0\), the measure \(\mu((a,b])=L_b^0-L_a^0\) on \([0,1]\) is finite, is supported on \(Z\), and has total mass \(L_1^0>0\) almost surely. Its two-time moment formula is \(\mathbb{E}[dL_s^0\,dL_t^0]=p_s(0)p_{t-s}(0)\,ds\,dt\) on \(0<s<t\), where \(p_u(0)=(2\pi u)^{-1/2}\); the corresponding formula on \(0<t<s\) follows by symmetry.

For a finite Borel measure \(\mu\) on \([0,1]\), its \(s\)-energy, for \(s>0\), is \(I_s(\mu)=\int\int |x-y|^{-s}\,\mu(dx)\mu(dy)\), with the kernel assigned value \(+\infty\) on the diagonal. The next criterion supplies the needed lower bound.

Energy criterion: If a finite nonzero Borel measure \(\mu\) is supported on \(A\subseteq\mathbb{R}\) and \(I_s(\mu)<\infty\) for some \(s>0\), then \(\dim_H(A)\ge s\).

Proof. Define the potential \(U(x)=\int |x-y|^{-s}\,\mu(dy)\). Since \(\int U(x)\,\mu(dx)=I_s(\mu)<\infty\), \(U(x)<\infty\) for \(\mu\)-almost every \(x\). In particular, \(\mu\) has no atoms: an atom of mass \(a>0\) at \(x\) would contribute \(a^2\cdot\infty\) to the energy. Choose \(K<\infty\) so that the Borel set \(E=\{x:U(x)\le K\}\) has positive \(\mu\)-measure; such a \(K\) exists since \(U\) is finite almost everywhere and \(\mu\) is nonzero. As \(\mu\) is supported on \(A\), we may take \(E\subseteq A\) up to a \(\mu\)-null set.

Consider any countable cover \((U_i)\) of \(E\) by arbitrary sets of diameter at most \(\delta\). For each set with \(E\cap U_i\ne\varnothing\), choose \(x_i\in E\cap U_i\), and write \(d_i=\operatorname{diam}(U_i)\). If \(d_i>0\), then \(E\cap U_i\subseteq[x_i-d_i,x_i+d_i]\). Since \(|x_i-y|\le d_i\) throughout that interval, the potential bound implies \(\mu([x_i-d_i,x_i+d_i])\le Kd_i^s\). Thus the outer measure \(\mu^*(E\cap U_i)\le Kd_i^s\). If \(d_i=0\), the nonempty set \(U_i\) is a singleton, and the absence of atoms gives \(\mu^*(E\cap U_i)=0\). Countable subadditivity of outer measure now yields

0<\mu(E)\le\sum_i\mu^*(E\cap U_i)\le K\sum_i d_i^s.

Taking the infimum over covers shows \(\mathcal{H}^s_\delta(E)\ge\mu(E)/K>0\) for every \(\delta>0\). Hence \(\mathcal{H}^s(E)>0\), so \(\dim_H(A)\ge\dim_H(E)\ge s\). \(\square\)

We apply this criterion to the local-time measure. For \(0<s<\tfrac12\), the stated moment formula and nonnegative integration give

\begin{aligned} \mathbb{E}I_s(\mu) &=2\int_{0<s<t<1}(t-s)^{-s}p_s(0)p_{t-s}(0)\,ds\,dt\\ &=\frac1\pi\int_{\substack{u>0,\ v>0\\u+v<1}}u^{-1/2}v^{-s-1/2}\,du\,dv\\ &\le\frac{2}{\pi}\int_0^1v^{-s-1/2}\,dv =\frac{2}{\pi(1/2-s)}<\infty. \end{aligned}

Here \(u\) is the earlier time and \(v\) the time gap. For fixed \(v\), the integral of \(u^{-1/2}\) from \(0\) to \(1-v\) is \(2\sqrt{1-v}\le2\); the final integral is finite exactly when \(s<\tfrac12\). Finite expected energy implies \(I_s(\mu)<\infty\) almost surely. The measure has positive mass and is supported on \(Z\) almost surely, so the energy criterion gives \(\dim_H(Z)\ge s\) almost surely. Taking the countable sequence \(s=\tfrac12-1/j\), \(j\ge3\), gives \(\dim_H(Z)\ge\tfrac12\) almost surely.

Worked Example: Checking the Energy Estimate at \(s=\tfrac14\)

Substitute \(s=\tfrac14\) into the bound just derived. The denominator is \(\tfrac12-\tfrac14=\tfrac14\), so

\mathbb{E}I_{1/4}(\mu)\le\frac{2}{\pi(1/4)}=\frac{8}{\pi}<\infty.

Thus \(I_{1/4}(\mu)\) is finite almost surely, and the energy criterion gives \(\dim_H(Z)\ge\tfrac14\) almost surely. This particular choice illustrates the method; using every rational \(s\) strictly between zero and one half gives the sharp lower bound.

Worked Example: Why Zero-Diameter Cover Sets Need Separate Treatment

Suppose a measure has finite \(s\)-energy for \(s>0\). It cannot put positive mass on a singleton \(\{x\}\): if \(\mu(\{x\})=a>0\), the part of the energy integral over \((x,x)\) contributes \(a^2|x-x|^{-s}=+\infty\). In the energy proof, a zero-diameter cover set is either empty or a singleton, so its intersection with the good set has outer measure zero.

For a positive-diameter cover set \(U_i\), one instead chooses \(x_i\in E\cap U_i\) and uses the containing interval of radius \(d_i=\operatorname{diam}(U_i)\). The potential bound gives \(\mu([x_i-d_i,x_i+d_i])\le Kd_i^s\). These two cases are both needed to justify the cover estimate, including when cover sets are not measurable.

Conclusion and a Common Pitfall

Theorem: For standard Brownian motion started at zero, the zero set \(Z=\{t\in[0,1]:B_t=0\}\) has \(\dim_H(Z)=\tfrac12\) almost surely.

The upper bound comes from the fact that an interval beginning at time \(k\delta\) has a probability of meeting the zero set that decreases on the order of \(k^{-1/2}\). The resulting expected number of intervals at mesh \(\delta\) is of order \(\delta^{-1/2}\). The lower bound is not a consequence of that count: it uses a nonzero measure carried by the zero set, namely local time, whose energy is finite for every exponent below one half.

Common mistake. Writing \(\mu(E\cap U_i)\) for arbitrary Hausdorff-cover sets \(U_i\) is unjustified: such a set need not be measurable, so that quantity may not be defined. The correction is to use outer measure \(\mu^*\), select a point \(x_i\in E\cap U_i\), and bound the outer measure by the measure of the containing interval centered at \(x_i\). Zero-diameter sets must be handled separately using the fact that finite positive-order energy precludes atoms.

Another tempting but insufficient argument is to cover all of \([0,1]\) by \(2^n\) dyadic intervals and infer a dimension bound from that count. That is only the trivial ambient cover; it does not use which intervals meet \(Z\). The decay in the probability of a hit is essential to obtaining the exponent one half.

Check Your Understanding

Use the definitions and estimates above to answer the following questions.

Why does the interval \([0,2^{-n}]\) contribute exactly one to \(N_n\) for every \(n\ge1\)?
In the upper-bound proof, why is it useful to show that \(\sum_n N_n2^{-ns}\) has finite expectation rather than only bounding \(\mathbb{E}N_n\) at each scale separately?
For which values of \(s>0\) does the displayed local-time energy bound remain finite, and where does that restriction enter the integral?
Why does finite \(s\)-energy imply that the measure has no atoms?
In the energy criterion, why can one not directly use \(\mu(E\cap U_i)\) for every set in a Hausdorff cover?

Hausdorff Dimension of the Brownian Zero Set

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