# Hausdorff Measure and Capacity

Definition 1 Let ${E \subset {\mathbb R}^n, 0 \le d < \infty, 0 < \delta \le \infty}$. Define

$\displaystyle \mathcal{H}^d_\delta(E) : = \inf \bigg\lbrace (1/2)^d \alpha(d) \sum_{i=1}^\infty diam(A_i)^d | E\subset \bigcup_{i=1}^\infty A_i, diam(A_i)\le \delta \bigg\rbrace,$

where ${\alpha(d) := \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}}$.

$\displaystyle \mathcal{H}^d (E) := \lim_{\delta \rightarrow 0} \mathcal{H}^d_\delta(E) = \sup_{\delta >0} \mathcal{H}^d_\delta(E)$

Then ${\mathcal{H}^d}$ is called d-dimensional Hausdorff measure in ${{\mathbb R}^n}$.

Remark 1 Hausdorff measure is a Borel outer measure, it is not Radon for ${0 \le d < n}$ because ${{\mathbb R}^n}$ is not ${\sigma-}$finite with respect to ${\mathcal{H}^d}$. And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set ${E}$ satisfies Caratheodory condition, if for any set ${A \subset {\mathbb R}^n}$, ${\mathcal{L}^n(A) = \mathcal{L}^n(A \cap E) + \mathcal{L}^n(A \setminus E)}$). Moreover, note that ${\alpha(d)}$ gives the volume of unit ball in ${d}$ dimension if ${d}$ is an integer, we naturally have (not trivially) ${\mathcal{H}^n = \mathcal{L}^n }$, where ${\mathcal{L}^n}$ is the n-dimensional Lebesgue measure.

Here are some basic properties of Hausdorff measures:

Theorem 2 ${\mathcal{H}_\delta^1 =\mathcal{L}^1}$ on ${{\mathbb R}^1}$, for all ${0< \delta \le \infty}$. And hence ${\mathcal{H}^1 =\mathcal{L}^1}$ on ${{\mathbb R}^1}$.

Proof: Note that ${\alpha(1) = 2}$, we have ${(1/2)^d \alpha(d) = 1}$ for ${d = 1}$.

$\displaystyle \mathcal{L}^1(E) : = \inf \bigg\lbrace \sum_{i=1}^\infty diam(A_i) | E\subset \bigcup_{i=1}^\infty A_i \bigg\rbrace$

$\displaystyle \le \inf \bigg\lbrace \sum_{i=1}^\infty diam(A_i) | E\subset \bigcup_{i=1}^\infty A_i, diam(A_i)\le \delta \bigg\rbrace = \mathcal{H}^1_\delta(E)$

Denote by ${I_k:= [k\delta, k\delta+\delta)}$ for ${k \in {\mathbb Z}}$. Then

$\displaystyle \mathcal{L}^1(E) : = \inf \bigg\lbrace \sum_{i=1}^\infty diam(A_i) | E\subset \bigcup_{i=1}^\infty A_i \bigg\rbrace$

$\displaystyle \ge \inf \bigg\lbrace \sum_{i=1}^\infty \sum_{k=-\infty}^\infty diam(A_i \cap I_k) | E\subset \bigcup_{i=1}^\infty A_i\bigg\rbrace \ge \mathcal{H}^1_\delta(E).$

So ${\mathcal{H}^1_\delta(E) = \mathcal{L}^1(E)}$ for all ${\delta> 0}$, hence ${\mathcal{H}^1 =\mathcal{L}^1}$ on ${{\mathbb R}^1}$. $\Box$

Theorem 3 ${\mathcal{H}^s \equiv 0}$ on ${{\mathbb R}^n}$ for ${s > n}$.

Proof: Let ${E}$ be a unit cube in ${{\mathbb R}^n}$, we can divide it into ${m^n}$ cubes with side-length ${\frac{1}{m}}$, diameter ${\frac{\sqrt{n}}{m}}$. So

$\displaystyle \mathcal{H}^s_{\frac{\sqrt{n}}{m}}(E) \le \alpha(s) m^n \left( \frac{\sqrt{n}}{2m} \right)^s = \frac{\alpha(s) n^{s/2}}{2^s} m^{n-s}.$

Send ${m \rightarrow \infty}$ we will have ${\mathcal{H}^s(E) = 0}$, hence ${\mathcal{H}^s \equiv 0}$ on ${{\mathbb R}^n}$. $\Box$

Theorem 4 Let ${E \subset {\mathbb R}^n}$, ${0 \le s < t <\infty}$, then

1. If ${\mathcal{H}^s(E) < +\infty}$, then ${\mathcal{H}^t(E) = 0}$.
2. If ${\mathcal{H}^t(E) > 0}$, then ${\mathcal{H}^s(E) = + \infty}$.

Proof: Note that the second statement is the contraposition of the first one. To prove the first statement, fix a ${\delta > 0}$, there exists a cover of ${E}$, ${A_i}$ with ${diam(A_i) < \delta}$, such that

$\displaystyle \sum_{i=1}^\infty \alpha(s) \left( \frac{diam(A_i)}{2} \right)^s < \mathcal{H}^s(E) + 1.$

Then we have

$\displaystyle \mathcal{H}^t_\delta(E) \le \sum_{i=1}^\infty \alpha(t) \left( \frac{diam(A_i)}{2} \right)^t \le (\delta/2)^{t-s} \frac{\alpha(t)}{\alpha(s)} \sum_{i=1}^\infty \alpha(s) \left( \frac{diam(A_i)}{2} \right)^s$

$\displaystyle < (\delta/2)^{t-s} \frac{\alpha(t)}{\alpha(s)}(\mathcal{H}^s(E) + 1).$

Send ${\delta \rightarrow 0^+}$ we will have ${\mathcal{H}^t(E) = 0}$. $\Box$

Having this theorem, we know that for any set ${E \subset {\mathbb R}^n}$ there exists only one ${s \ge 0}$ such that ${\mathcal{H}^s(E)}$ is neither ${0}$ nor ${\infty}$. It is natural to define this ${s}$ to be its dimension.

Definition 5 We say the Hausdorff dimension of a set ${E \subset {\mathbb R}^n}$ is

$\displaystyle H_{dim} (E) := \inf \{ 0 \le s < \infty | \mathcal{H}^s(E) = 0\}.$

Note that the Hausdorff dimension need not be an integer. Even if ${H_{dim}(E) = s}$ is an integer, it doesn’t mean ${E}$ is a ${s}$-dimensional surface in any sense.

Another important and related measure is called capacity, and it is defined as follows.

Definition 6 Let ${E \subset {\mathbb R}^n}$, we define the p-capacity of ${E}$ by

$\displaystyle \Gamma_p(E) := \inf \left\{ \int_{R^n} | \nabla f|^p dx: f \ge 0, f \in C_c^\infty({\mathbb R}^n) , E \subset Int\{f \ge 1\} \right\}.$

Obviously, by density we can consider ${f \in A^p:= \{ f : {\mathbb R}^n \rightarrow {\mathbb R} | Df \in L^p({\mathbb R}^n; {\mathbb R}^n), f \in L^{\frac{np}{n-p}}({\mathbb R}^n) \}}$ instead of ${f \in C_c^\infty({\mathbb R}^n)}$.

It’s not hard to see ${\Gamma_p}$ is an outer measure, by know that given a sequence of functions ${f_n \in C_c^\infty}$, denote by ${w(x) := \sup_n f_n}$ and ${\lambda(x) := \sup_n |\nabla f_n|}$, then ${|w(x)| \le \lambda(x)}$ ${\mathcal{L}^n}$-a.e. As we know, given an outer measure, one can construct a ${\sigma}$-algebra by Caratheodory condition such that it is a measure restricted to this ${\sigma}$-algebra. But in this case, we are not that interested in it because the ${\sigma}$-algebra only contains sets of capacity 0 or ${\infty}$. And here are some properties of capacities that one can easily prove.

Theorem 7 (Properties of ${\Gamma_p}$)

1. ${\Gamma_p(E) = \inf \{\Gamma_p(U) : U}$ open, ${E \subset U \}.}$
2. ${\Gamma_p(\lambda E) = \lambda^{n-p} \Gamma_p(E)}$, for ${\lambda > 0}$.
3. ${\Gamma_p(E) \le C(n,p) \mathcal{H}^{n-p}(E)}$.
4. ${\mathcal{L}^n (E) \le C(n,p) \Gamma_p(E)^{\frac{n}{n-p}}}$.

From the scaling, we can see that the capacity does not measure the “volume” of the set. And according this scaling, it is natural to look for the relations between ${\Gamma_p}$ and ${\mathcal{H}^{n-p}}$.

Theorem 8 Assume ${1< p < n}$, if ${\mathcal{H}^{n-p} (E) < \infty}$, then ${\Gamma_p(E) = 0}$.

Note that this is a refinement of the third statement of last theorem for \$latex {1

Proof: First we claim that there exists a constant ${C = C(n,E)}$ such that for any neighborhood ${V}$ containing ${E}$, there exists an open set ${W}$ and ${f \in A^p}$ such that

$\displaystyle E \subset W \subset \{ f = 1\},~ spt(f) \subset V,~ \int_{{\mathbb R}^n} |Df|^p \, dx \le C(n,E).$

To proof the claim, let ${V}$ be a neighborhood of ${E}$ and set ${\delta := \frac{1}{2} dist(E, {\mathbb R}^n - V)}$. Since ${\mathcal{H}^{n-p}(E) < \infty}$, we can find a sequence of ball ${\{B(b_i,r_i)\}}$ with ${2r_i < \delta}$ such that ${E \subset \cup_{i \in {\mathbb N}} B(b_i, r_i)}$, and

$\displaystyle \sum_{i \in {\mathbb N}} \alpha(n-p) r_i^{n-p} < \mathcal{H}^{n-p}(E) + 1.$

Set

$\displaystyle h_i(x) = 1, \text{ if } |x - b_i| < r_i , \quad h_i(x) = 0 \text{ if } |x - b_i| \ge 2 r_i,$

$\displaystyle h_i(x) = 2 - \frac{|x - b_i|}{r_i}, \text{ if } r_i \le |x - b_i| < 2 r_i.$

Then ${\int_{{\mathbb R}^n} |\nabla h_i|^p \, dx = C r_i^{n-p}}$. Set ${f(x) := \sup_i h_i(x)}$ and ${W := \cup_{i \in {\mathbb N}} B(b_i, r_i)}$, then ${f \in A^p, spt(f) \subset V, W \subset \{f = 1\}}$ and

$\displaystyle \int_{{\mathbb R}^n} |D f|^p \, dx \le \sum_{i \in {\mathbb N}} \int_{{\mathbb R}^n} |D h_i|^p \le C \sum_{i \in {\mathbb N}} r_i^{n-p} \le C(n,E).$

Now we use the claim inductively. We find a sequence of ${f_k \in A^p}$, ${V_k}$ from the claim, and relate them by ${\bar{V}_{k+1} \subset Int \{ f_k = 1\}}$. Set ${S_j = \sum_{k=1}^j \frac{1}{k}}$ and ${g_j:= \frac{1}{S_j} \sum_{k= 1}^j \frac{f_k}{k}}$. Then ${g_j \in A^p}$ and ${g_j = 1}$ on ${V_{j+1}}$. Therefore

$\displaystyle \Gamma_p(E) \le \int_{{\mathbb R}^n} |D g_j|^p \, dx \le \frac{1}{S_j^p} \sum_{k = 1}^j \frac{1}{k^p} \int_{{\mathbb R}^n} |D f_k|^p \, dx \le \frac{C}{S_j^p} \sum_{k = 1}^j \frac{1}{k^p} \rightarrow 0,$

since ${spt(Df_k)}$ are discrete and ${p > 1}$. $\Box$

The next important relation is that, given ${\Gamma_p(E) = 0}$, we will have some information on its Hausdorff dimension. Before proving it, we need a Lemma.

Lemma 9 Let ${f \in L_{loc}^1 ({\mathbb R}^n)}$, suppose ${0 \le s < n}$ and define

$\displaystyle \Lambda_s := \left\{ x \in {\mathbb R}^n: \limsup_{r \rightarrow 0} \frac{1}{r^s} \int_{B(x,r)} |f| > 0 \right\}.$

Then ${\mathcal{H}^s(\Lambda_s) = 0}$.

Proof: From Lebesgue differential theorem one can see ${\mathcal{L}^n(\Lambda_s) = 0}$. Then by the absolute continuity of Lebesgue integral, there exists a ${d>0}$ and ${C}$ such that ${\int_U |f| < \gamma}$ whenever ${\mathcal{L}^n (U) < d}$. Fix a ${\delta > 0}$ and an ${\varepsilon > 0}$, we define a set

$\displaystyle \Lambda_s^\varepsilon := \left\{ x \in {\mathbb R}^n: \limsup_{r \rightarrow 0} \frac{1}{r^s} \int_{B(x,r)} |f| > \varepsilon \right\}.$

Since ${\mathcal{L}^n(\Lambda_s^\varepsilon) = 0}$, we can fix an open set ${U \supset \Lambda_s^\varepsilon}$ with ${\mathcal{L}^n(U) < d}$, and consider a family of balls

$\displaystyle \mathcal{B} := \{ B(x,r) : x \in \Lambda_s^\varepsilon , r < \delta , B(x,r) \subset U, \int_{B(x,r)} |f| > \frac{\varepsilon}{2} r^s \}.$

Then this family of balls cover ${\Lambda_s^\varepsilon}$, and by Vitali’s covering theorem, there exists a sequence of disjoint balls ${B_i}$ such that ${\Lambda_s^\varepsilon \subset \cup_{i \in {\mathbb N}} 5B_i}$. Hence

$\displaystyle \mathcal{H}^s_{10\delta}(\Lambda_s^\varepsilon) \le C\sum_{i \in {\mathbb N}} r_i^s \le \frac{C}{\varepsilon} \int_{U} |f| \le \frac{C \gamma}{\varepsilon}.$

Then we can take ${\delta \rightarrow 0, \gamma \rightarrow 0, \varepsilon \rightarrow 0}$ to yield the result. $\Box$

Theorem 10 Let ${E \subset {\mathbb R}^n}$ and ${1 \le p < \infty}$. If ${\Gamma_p(E) = 0}$, then ${\mathcal{H}^s(E) = 0}$ for all ${s > n - p}$.

Proof: Since ${\Gamma_p(E) = 0}$, we can choose a sequence of functions ${f_i \in A^p, E \subset int\{ f_i \ge 1\}}$ such that ${\int_{{\mathbb R}^n} |Df_i| < \frac{1}{2^i}}$. Define ${g := \sum_{i \in {\mathbb N}} f_i}$, by GNS inequality we know ${g}$ is well defined in ${A^p}$, and ${E \subset int \{ g \ge m\}}$ for any ${m \ge 1}$. So ${(g)_{x,r}}$, the average of ${g}$ on ${B(x.r)}$, goes to 0 as ${r \rightarrow 0}$ if ${x \in E}$. Now I claim that

$\displaystyle E \subset \left\{ x \in {\mathbb R}^n: \limsup_{r \rightarrow 0} \frac{1}{r^s} \int_{B(x,r)} |Dg|^p = +\infty \right\}.$

Assume by contradiction, there exists an ${x \in E}$ such that ${ \limsup_{r \rightarrow 0} \frac{1}{r^s} \int_{B(x,r)} |Dg|^p < M}$, by Poincare inequality, we know that

$\displaystyle \int_{B(x,r)} |g - (g)_{x,r}|^p \le r^p \int_{B(x,r)} |Dg|^p < 2M r^{p+s},$

for ${r}$ small enough. Then

$\displaystyle |(g)_{x,r/2} - (g)_{x,r}| \le \left( \frac{C}{r^n} \int_{B(x,r/2)} |g - (g)_{x,r}|^p \right)^{\frac{1}{p}} \le C r^{\frac{p-n+s}{p}} = C r^\theta,$

for some ${\theta = p-n+s > 0}$. This means ${(g)_{x,r}}$ converges as ${r \rightarrow 0}$, which leads to a contradiction. Therefore, ${E \subset \Lambda_s}$, and by the previous lemma we know ${\mathcal{H}^s(\Lambda_s) = 0}$ since ${|Dg| \in L^1_{loc}({\mathbb R}^n)}$. $\Box$

For ${p = 1}$ case, we can say more about the relation between capacity and Hausdorff measure since we will have more geometric insights. For instance, we can take advantage of isoperimetric inequality and co-area formula to prove the following result.

Theorem 11 Let ${E \subset {\mathbb R}^n}$, then ${\Gamma_1(E) = 0}$ if and only if ${\mathcal{H}^{n-1}(E) = 0}$.

Proof: From Theorem 8 we know that ${\mathcal{H}^{n-1}(E) = 0}$ implies ${\Gamma_1(E) = 0}$. Assume ${\Gamma_1(E) = 0}$, for any ${\varepsilon > 0}$, we can find an ${f \in C_c^\infty({\mathbb R}^n)}$ such that ${E \subset int\{ f \ge 1\}}$ and ${\int_{{\mathbb R}^n} |Df| < \varepsilon}$. By co-area formula, we know

$\displaystyle \int_{{\mathbb R}^n} |Df| \ge \int_0^1 \mathcal{H}^{n-1} (\partial \{f \ge t\}) \,dt.$

This implies ${\mathcal{H}^{n-1} (\partial \{f \ge t\}) \le \int_{{\mathbb R}^n} |Df|}$ and ${\mathcal{L}^n(\{f \ge t\}) > 0}$ for some ${t \in (0,1)}$. Fix this ${t}$, by isoperimetric inequality, one can see ${\mathcal{L}^n(\{f \ge t\}) \le C \mathcal{H}^{n-1} (\partial \{f \ge t\})^{\frac{n}{n-1}} < \infty}$. Obviously we know ${E \subset \{f \ge t\}}$, then for each ${x \in E}$, we can find an ${r>0}$ such that

$\displaystyle \frac{\mathcal{L}^n(\{f \ge t\} \cap B(x,r))}{\alpha(n)r^n} = \frac{1}{4}.$

Since ${\mathcal{L}^n(\{f \ge t\} \cap B(x,r))}$ is a small portion in ${B(x,r)}$, by relative isoperimetric inequality we have

$\displaystyle C r^{n-1} = \mathcal{L}^n(\{f \ge t\} \cap B(x,r))^{\frac{n-1}{n}} \le \mathcal{H}^{n-1}(\partial \{f \ge t\} \cap B(x,r)).$

Since these balls cover ${E}$, by Vitali’s covering theorem, there exists a sequence of disjoint balls ${B(x_i, r_i)}$ such that ${E \subset \cup_{i \in {\mathbb N}} B(x_i , 5r_i)}$. From the estimates we get above,

$\displaystyle \sum_{i \in {\mathbb N}} (5r_i)^{n-1} \le C \mathcal{H}^{n-1}(\partial \{f \ge t\}) \le \int_{{\mathbb R}^n} |Df| < \varepsilon.$

And we know that for each ${i, r_i < C\varepsilon^{\frac{1}{n-1}}}$. Hence ${\mathcal{H}^{n-1}_{C\varepsilon^{1/(n-1)}} (E) \le C\varepsilon}$. Send ${\varepsilon \rightarrow 0}$ we will get the result. $\Box$

Remark 2 In the previous theorem, we actually prove the smallness of ${\Gamma_p}$ is equivalent to the smallest of ${\mathcal{H}^{n-1}_\infty}$, which we don’t have for ${p > 1}$.

The next important result is fine property of Sobolev functions. Comparing to Lebesgue integrable functions, Sobolev functions have a smaller singular set, and their representatives enjoy a better continuity result. To show it, we first need a Chebyshev type inequality.

Lemma 12 Let ${f \in A^p}$ and ${\varepsilon > 0}$, define the set

$\displaystyle E_\varepsilon := \{ x \in {\mathbb R}^n: (f)_{x,r} > \varepsilon , \text{ for some } r > 0 \}.$

Then

$\displaystyle \Gamma_p(E_\varepsilon) \le \frac{C}{\varepsilon^p} \int_{{\mathbb R}^n} |Df|^p \, dx.$

Proof: First note that, without loss of generality, we can assume ${\varepsilon = 1}$ by homogeneity. For each ${x \in E_1}$, we consider a ball ${B(x,r)}$ such that ${(f)_{x,r} > 1}$. This collection of balls cover ${E_1}$, and by Besicovitch covering theorem, there exist ${C_n \in {\mathbb N}}$ and countable collections of disjoint balls ${\mathcal{B}_1 , \cdots \mathcal{B}_{C_n}}$, such that

$\displaystyle E_1 \subset \bigcup_{i =1}^{C_n} \bigcup_{B \in \mathcal{B}_i} B, \quad \text{ and } \quad (f)_B > 1 \text{ for any } B \in \bigcup_{i =1}^{C_n} \mathcal{B}_i.$

Denote by ${\{Q_i^j\}_{j \in {\mathbb N}}}$ the element in ${\mathcal{B}_i}$ for ${i = 1 , 2 , \cdots , C_n}$. For each ${i,j}$, consider the function ${((f)_{Q_i^j} - f)^+}$, which is in ${A^p}$, on ${Q_i^j}$. By extension theorem we know this function can be extended to ${{\mathbb R}^n}$, say ${h_i^j}$, such that

$\displaystyle \int_{{\mathbb R}^n} |Dh_i^j|^p \, dx \le C \int_{Q_i^j} |Df|^p dx,$

for all ${i,j}$, where ${C}$ depends only on ${n,p}$. Then we have ${f + h_i^j \ge (f)_{Q_i^j} \ge 1}$ in ${Q_i^j}$ for all ${i,j}$. Define ${h := \sup_{i,j} h_i^j \in A^p}$, then ${E_1 \subset int \{f + h \ge 1\}}$ since ${E_1}$ is open. Therefore

$\displaystyle \Gamma_p(E_1) \le \int_{{\mathbb R}^n}|D(f+h)|^p \, dx \le C \left( \int_{{\mathbb R}^n}|Df|^p \, dx + \sum_{i = 1}^{C_n} \sum_{Q_i^j \in \mathcal{B}_i} \int_{Q_i^j} |Df|^p \, dx \right)$

$\displaystyle \le C \int_{{\mathbb R}^n}|Df|^p \, dx.$

$\Box$

Now we are ready to prove the fine properties of Sobolev functions.

Theorem 13 Let ${f \in W^{1,p}({\mathbb R}^n), 1 \le p < n}$. Then there exists a Borel set ${E}$ such that ${\Gamma_p(E) = 0}$, and ${\lim_{r \rightarrow 0}(f)_{x,r} =: f^*(x)}$ exists for any ${x \in {\mathbb R}^n \setminus E}$. In addition,

$\displaystyle \lim_{r\rightarrow 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f - f^*(x)|^{\frac{np}{n-p}} \, dy = 0$

for all ${x \in {\mathbb R}^n \setminus E}$. Furthermore, given any ${\varepsilon > 0}$, there exists an open set ${U}$ such that ${f^*}$ is continuous restricted on ${{\mathbb R}^n \setminus U}$.

Proof: For each ${i \in {\mathbb N}}$, choose an ${f_i \in W{1,p}({\mathbb R}^n) \cap C^\infty({\mathbb R}^n)}$ such that ${\int_{{\mathbb R}^n} |Df - Df_i|^p \le \frac{1}{2^{(p+1)i}}}$. Define the set

$\displaystyle B_i:= \left\{ x \in {\mathbb R}^n: \frac{1}{|B(x,r)|} \int_{B(x,r)} |f - f_i| \, dy > \frac{1}{2^i}, \text{ for some }r > 0 \right\}.$

From Lemma 14, we know ${\Gamma_p(B_i) \le 2^{pi} \frac{1}{2^{(p+1)i}} = \frac{1}{2^i}}$. Consider the set

$\displaystyle A:= \left\{ x \in {\mathbb R}^n: \limsup_{r \rightarrow 0} \frac{1}{r^{n-p}} \int_{B(x,r)} |Df|^p \, dx > 0 \right\},$

then by Theorem 9 and Lemma 10, we have ${\mathcal{H}^{n-p}(A) = \Gamma_p(A) = 0}$. For any ${x \in {\mathbb R}^n \setminus A}$, by Poincare’s inequality we have

$\displaystyle \lim_{r \rightarrow 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f - (f)_{x,r}|^{\frac{np}{n-p}} \, dy = 0.$

Then for any ${x \in {\mathbb R}^n \setminus (A \cup B_i)}$, we have

$\displaystyle \limsup_{r \rightarrow 0} |f_i(x) - (f)_{x,r}|$

$\displaystyle \le \limsup_{r \rightarrow 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f - f_i| + |f - (f)_{x,r}| + |f_i - f_i(x)| \, dy \le \frac{1}{2^i}.$

Define ${E_k = A \cup (\cup_{i = k}^\infty B_i)}$, then for ${i,j \ge k, x \in {\mathbb R}^n \setminus E_k}$, we have ${|f_i(x) - f_j(x)| \le \frac{1}{2^i} + \frac{1}{2^j}}$. This implies ${f_i}$ converges uniformly to a continuous function ${g}$ in ${{\mathbb R}^n \setminus E_k}$. So

$\displaystyle |g - (f)_{x,r}| \le |f_i - g| + |f_i - (f)_{x,r}| \rightarrow 0,$

as ${i \rightarrow \infty, r \rightarrow 0}$, which implies ${g(x) = \lim_{r \rightarrow 0} (f)_{x,r} = f^*(x)}$. Define ${E := \cap_{k \in {\mathbb N}} E_k}$, then

$\displaystyle \Gamma_p(E) \le \limsup_{k \rightarrow \infty} \Gamma_p(E_k)$

$\displaystyle \le \limsup_{k \rightarrow \infty} (\Gamma_p(A) + \sum_{i = k}^\infty \Gamma_p(B_i)) = \limsup_{k \rightarrow \infty} \frac{1}{2^{k -1}}= 0.$

And we know that ${\lim_{r \rightarrow 0} (f)_{x,r}}$ exists for all ${x \in {\mathbb R}^n \setminus E}$. In addition,

$\displaystyle \frac{1}{|B(x,r)|} \int_{B(x,r)} |f - f^*(x)|^{\frac{np}{n-p}} \, dy$

$\displaystyle \le |f^*(x) - (f)_{x,r}|^{\frac{np}{n-p}} + \frac{1}{|B(x,r)|} \int_{B(x,r)} |f - (f)_{x,r}|^{\frac{np}{n-p}} \, dy \rightarrow 0.$

For the continuity statement, given any ${\varepsilon > 0}$, we can find an ${E_k}$ such that ${\Gamma_p(E_k) < \frac{\varepsilon}{2}}$. From theorem 8, we can find an open set ${U \supset E_k}$ such that ${\Gamma_p(U) < \varepsilon}$. Then we are done since we know ${f^*}$ is continuous restricted on ${{\mathbb R}^n \setminus E_k}$. $\Box$