Solution.
Let m m m R \R R m m m μ \mu μ R = A ∪ A c \R = A \cup A^\comp R = A ∪ A c μ ( A ) = 0 \mu\p{A} = 0 μ ( A ) = 0 m ( A c ) = 0 m\p{A^\comp} = 0 m ( A c ) = 0 A A A m ( A c ) = 0 m\p{A^\comp} = 0 m ( A c ) = 0 x ∈ A x \in A x ∈ A ε > 0 \epsilon > 0 ε > 0
E k = { x ∈ A | lim sup δ → 0 μ ( [ x − δ , x + δ ] ) m ( [ x − δ , x + δ ] ) > 1 k } . E_k = \set{x \in A \st \limsup_{\delta\to0}\,\frac{\mu\p{\br{x - \delta, x + \delta}}}{m\p{\br{x - \delta, x + \delta}}} > \frac{1}{k}}. E k = { x ∈ A ∣ ∣ δ → 0 lim sup m ( [ x − δ , x + δ ] ) μ ( [ x − δ , x + δ ] ) > k 1 } . Since μ \mu μ R \R R U ⊇ A U \supseteq A U ⊇ A μ ( U ∖ A ) < ε \mu\p{U \setminus A} < \epsilon μ ( U ∖ A ) < ε A A A
μ ( U ∖ A ) = μ ( U ) − μ ( A ) = μ ( U ) ⟹ μ ( U ) < ε . \mu\p{U \setminus A} = \mu\p{U} - \mu\p{A} = \mu\p{U} \implies \mu\p{U} < \epsilon. μ ( U ∖ A ) = μ ( U ) − μ ( A ) = μ ( U ) ⟹ μ ( U ) < ε . Then for any x ∈ E k x \in E_k x ∈ E k δ x > 0 \delta_x > 0 δ x > 0 I ( x , δ x ) = [ x − δ x , x + δ x ] ⊆ U I\p{x, \delta_x} = \br{x - \delta_x, x + \delta_x} \subseteq U I ( x , δ x ) = [ x − δ x , x + δ x ] ⊆ U
μ ( I ( x , δ x ) ) m ( I ( x , δ x ) ) = μ ( [ x − δ x , x + δ x ] ) m ( [ x − δ x , x + δ x ] ) > 1 k . \frac{\mu\p{I\p{x, \delta_x}}}{m\p{I\p{x, \delta_x}}} = \frac{\mu\p{\br{x - \delta_x, x + \delta_x}}}{m\p{\br{x - \delta_x, x + \delta_x}}} > \frac{1}{k}. m ( I ( x , δ x ) ) μ ( I ( x , δ x ) ) = m ( [ x − δ x , x + δ x ] ) μ ( [ x − δ x , x + δ x ] ) > k 1 . We also have
E k ⊆ ⋃ x ∈ E k I ( x , δ x 5 ) . E_k \subseteq \bigcup_{x \in E_k} I\p{x, \frac{\delta_x}{5}}. E k ⊆ x ∈ E k ⋃ I ( x , 5 δ x ) . By the Vitali covering lemma, there exist a countable collection { x n } n ⊆ E k \set{x_n}_n \subseteq E_k { x n } n ⊆ E k
E k ⊆ ⋃ x ∈ E k I ( x , δ x 5 ) ⊆ ⋃ n = 1 ∞ I ( x i , δ i ) , E_k \subseteq \bigcup_{x \in E_k} I\p{x, \frac{\delta_x}{5}} \subseteq \bigcup_{n=1}^\infty I\p{x_i, \delta_i}, E k ⊆ x ∈ E k ⋃ I ( x , 5 δ x ) ⊆ n = 1 ⋃ ∞ I ( x i , δ i ) , where δ i = δ x i \delta_i = \delta_{x_i} δ i = δ x i
m ( E k ) ≤ m ( ⋃ i = 1 ∞ I ( x i , δ i ) ) ≤ ∑ i = 1 ∞ m ( I ( x i , δ i ) ) ≤ ∑ i = 1 ∞ k μ ( I ( x i , δ i ) ) ≤ k μ ( ⋃ i = 1 ∞ I ( x i , δ i ) ) ≤ k μ ( U ) < k ε . \begin{aligned}
m\p{E_k}
\leq m\p{\bigcup_{i=1}^\infty I\p{x_i, \delta_i}}
&\leq \sum_{i=1}^\infty m\p{I\p{x_i, \delta_i}} \\
&\leq \sum_{i=1}^\infty k\mu\p{I\p{x_i, \delta_i}} \\
&\leq k\mu\p{\bigcup_{i=1}^\infty I\p{x_i, \delta_i}} \\
&\leq k\mu\p{U} \\
&< k\epsilon.
\end{aligned} m ( E k ) ≤ m ( i = 1 ⋃ ∞ I ( x i , δ i ) ) ≤ i = 1 ∑ ∞ m ( I ( x i , δ i ) ) ≤ i = 1 ∑ ∞ k μ ( I ( x i , δ i ) ) ≤ k μ ( i = 1 ⋃ ∞ I ( x i , δ i ) ) ≤ k μ ( U ) < k ε . Since k k k ε \epsilon ε ε → 0 \epsilon \to 0 ε → 0 m ( E k ) = 0 m\p{E_k} = 0 m ( E k ) = 0
m ( ⋃ k = 1 ∞ E k ) ≤ ∑ k = 1 ∞ m ( E k ) = 0 , m\p{\bigcup_{k=1}^\infty E_k} \leq \sum_{k=1}^\infty m\p{E_k} = 0, m ( k = 1 ⋃ ∞ E k ) ≤ k = 1 ∑ ∞ m ( E k ) = 0 , and
x ∈ ( ⋃ k = 1 ∞ E k ) c ⟺ lim sup δ → 0 μ ( x − δ , x + δ ) 2 ε ≤ 1 k ∀ k ≥ 1 , x \in \p{\bigcup_{k=1}^\infty E_k}^\comp \iff \limsup_{\delta\to0}\,\frac{\mu\p{x - \delta, x + \delta}}{2\epsilon} \leq \frac{1}{k}\ \forall k \geq 1, x ∈ ( k = 1 ⋃ ∞ E k ) c ⟺ δ → 0 lim sup 2 ε μ ( x − δ , x + δ ) ≤ k 1 ∀ k ≥ 1 , i.e., the Lebesgue differentiate theorem holds Lebesgue almost everywhere.