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 knowing 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

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

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