Thu-20-11-2025 11:40
profe: Cecilia De Vita
status:
tags:
Ejercicio 1
Para cada n ∈ N , n\in \mathbb{N}, n ∈ N , f n : R → R f_{n}:\mathbb{R}\to \mathbb{R} f n : R → R E = { x ⊆ R : ( f n ( x ) ) n ∈ N converge } E=\{ x \subseteq \mathbb{R}:(f_{n}(x))_{n\in \mathbb{N}} \text{ converge} \} E = { x ⊆ R : ( f n ( x ) ) n ∈ N converge }
Sugerencia: Recordar que toda sucesión convergente es de Cauchy.
Idea: Tratar de escribir a E E E x ∈ R . x \in \mathbb{R}. x ∈ R . ( f n ) n ∈ N ⊆ R ( f_{n} )_{n \in \mathbb{N}}\subseteq \mathbb{R} ( f n ) n ∈ N ⊆ R ( f n ) n ∈ N ( f_{n} )_{n \in \mathbb{N}} ( f n ) n ∈ N ∀ E > 0 ∃ k 0 ∈ N ∣ ∣ f n ( x ) − f m ( x ) ∣ < E ∀ n , m ≥ k 0 \forall\mathcal{E}> 0\:\exists\:k_{0}\in \mathbb{N}\bigm||f_{n}(x)-f_{m}(x)|<\mathcal{E}\quad\forall n,m\geq k_{0} ∀ E > 0 ∃ k 0 ∈ N ∣ f n ( x ) − f m ( x ) ∣ < E ∀ n , m ≥ k 0
Equivalentemente:
∀ l ∈ N ∃ k ∈ N ∣ ∣ f n ( x ) − f m ( x ) ∣ < 1 l ∀ n , m ≥ k \forall \mathscr{l}\in \mathbb{N}\:\exists\:k\in \mathbb{N}\bigm| |f_{n}(x)-f_{m}(x)|< \frac{1}{\mathscr{l}}\quad\forall n,m\geq k ∀ l ∈ N ∃ k ∈ N ∣ f n ( x ) − f m ( x ) ∣ < l 1 ∀ n , m ≥ k Llamo E n , m , l = { x ∈ R : ∣ f n ( x ) − f m ( x ) ∣ < 1 l } E_{n,m,\mathscr{l}}=\left\{ x \in \mathbb{R} :|f_{n}(x)-f_{m}(x)|< \frac{1}{\mathscr{l}} \right\} E n , m , l = { x ∈ R : ∣ f n ( x ) − f m ( x ) ∣ < l 1 }
E = ⋂ l ∈ N ⋃ k ∈ N ⋂ ( n , m ) : n , m ≥ k E n , m , l E=\bigcap_{\mathscr{l}\in \mathbb{N}}\bigcup_{k \in \mathbb{N}}\bigcap_{(n,m):n,m\geq k}E_{n,m,\mathscr{l}} E = l ∈ N ⋂ k ∈ N ⋃ ( n , m ) : n , m ≥ k ⋂ E n , m , l Si pruebo que E m , n , l E_{m,n,\mathscr{l}} E m , n , l ∀ m , n , l ∈ N \forall m,n,\mathscr{l} \in \mathbb{N} ∀ m , n , l ∈ N E E E n , m , l ∈ N n,m,\mathscr{l}\in \mathbb{N} n , m , l ∈ N
∣ f n ( x ) − f m ( x ) ∣ < 1 l ⟺ − 1 l < ∣ f n ( x ) − f m ( x ) ∣ < 1 l ⟺ |f_{n}(x)-f_{m}(x)|< \frac{1}{\mathscr{l}}\iff-\frac{1}{\mathscr{l}}<|f_{n}(x)-f_{m}(x)|< \frac{1}{\mathscr{l}}\iff ∣ f n ( x ) − f m ( x ) ∣ < l 1 ⟺ − l 1 < ∣ f n ( x ) − f m ( x ) ∣ < l 1 ⟺ ⟺ f n ( x ) − f m ( x ) > − 1 l ∧ f n ( x ) − f m ( x ) < 1 l \iff f_{n}(x)-f_{m}(x)> - \frac{1}{\mathscr{l}} \land f_{n}(x)-f_{m}(x)< \frac{1}{\mathscr{l}} ⟺ f n ( x ) − f m ( x ) > − l 1 ∧ f n ( x ) − f m ( x ) < l 1 Luego
E m , n , l = { x ∈ R : f n ( x ) − f m ( x ) > − 1 l } ⏟ Medible pues f n , f m medibles. Entonces la resta es med. ∩ { x ∈ R : f n ( x ) − f m ( x ) < 1 l } ⏟ Medible por la misma raz o ˊ n. E_{m,n,\mathscr{l}}=\underbrace{ \left\{ x \in \mathbb{R}:f_{n}(x)-f_{m}(x)>- \frac{1}{\mathscr{l}} \right\} }_{ \text{Medible pues $f_{n},f_{m}$ medibles. Entonces la resta es med.} } \cap \underbrace{ \left\{ x \in \mathbb{R}:f_{n}(x)-f_{m}(x)< \frac{1}{\mathscr{l}} \right\} }_{ \text{Medible por la misma razón.} } E m , n , l = Medible pues f n , f m medibles. Entonces la resta es med. { x ∈ R : f n ( x ) − f m ( x ) > − l 1 } ∩ Medible por la misma raz o ˊ n. { x ∈ R : f n ( x ) − f m ( x ) < l 1 } Entonces, E m , n , l E_{m,n,\mathscr{l}} E m , n , l
Ejercicio 2
--00:30:00
Sea g : R → R g:\mathbb{R}\to \mathbb{R} g : R → R f : R → R f:\mathbb{R}\to \mathbb{R} f : R → R g ∘ f g\circ f g ∘ f
Sea a ∈ R a \in \mathbb{R} a ∈ R E a = { x ∈ R : g ∘ f ( x ) < a } E_{a}=\{ x \in \mathbb{R}:g \circ f (x)<a \} E a = { x ∈ R : g ∘ f ( x ) < a }
x ∈ E a ⟺ g ∘ f ( x ) < a ⟺ g ( f ( x ) ) < a ⟺ f ( x ) ∈ g − 1 ( − ∞ , a ) x \in E_{a}\iff g \circ f(x)<a \iff g(f(x))<a \iff f(x) \in g ^{-1}(-\infty,a) x ∈ E a ⟺ g ∘ f ( x ) < a ⟺ g ( f ( x )) < a ⟺ f ( x ) ∈ g − 1 ( − ∞ , a ) ⟺ x ∈ f − 1 ( g − 1 ( − ∞ , a ) ) \iff x \in f ^{-1}(g ^{-1}(-\infty,a)) ⟺ x ∈ f − 1 ( g − 1 ( − ∞ , a )) Entonces E a = f − 1 ( g − 1 ( ( − ∞ , a ) ) ) E_{a}=f ^{-1}(g ^{-1}((-\infty,a))) E a = f − 1 ( g − 1 (( − ∞ , a ))) g g g ( − ∞ , a ) (-\infty,a) ( − ∞ , a ) g − 1 ( ( − ∞ , a ) ) g ^{-1}((-\infty,a)) g − 1 (( − ∞ , a )) g − 1 ( − ∞ , a ) g ^{-1}(-\infty,a) g − 1 ( − ∞ , a ) ( I j ) j ∈ N (I_{j})_{j \in \mathbb{N}} ( I j ) j ∈ N
g − 1 ( ( − ∞ , a ) ) = ⋃ j = 1 ∞ I j g ^{-1}((-\infty,a))=\bigcup_{j=1}^{\infty} I_{j} g − 1 (( − ∞ , a )) = j = 1 ⋃ ∞ I j Entonces
E a = = f − 1 ( g − 1 ( − ∞ , a ) ) = f − 1 ( ⋃ j = 1 ∞ I j ) = Tarea ⋃ j = 1 ∞ f − 1 ( I j ) E_{a}==f ^{-1}(g ^{-1}(-\infty,a))=f ^{-1}\left( \bigcup_{j =1}^{\infty} I_{j} \right)\underset{ \text{Tarea} }{ = } \bigcup_{j=1}^{\infty} f ^{-1}(I_{j}) E a == f − 1 ( g − 1 ( − ∞ , a )) = f − 1 ( j = 1 ⋃ ∞ I j ) Tarea = j = 1 ⋃ ∞ f − 1 ( I j ) Si vemos que f − 1 ( I j ) f ^{-1}(I_{j}) f − 1 ( I j ) ∀ j ∈ N ⟹ E a \forall j \in \mathbb{N}\implies E_{a} ∀ j ∈ N ⟹ E a j ∈ N . j \in \mathbb{N}. j ∈ N . I j = ( a j , b j ) I_{j}=(a_{j},b_{j}) I j = ( a j , b j ) a j , b j ∈ R a_{j},b_{j} \in \mathbb{R} a j , b j ∈ R a j < b j . a_{j}<b_{j}. a j < b j .
f − 1 ( I j ) = { x ∈ R : f ( x ) ∈ ( a j , b j ) } = { x ∈ R : a j < f ( x ) < b j } = f ^{-1}(I_{j})=\{ x \in \mathbb{R}:f(x) \in(a_{j},b_{j}) \}=\{ x \in \mathbb{R}:a_{j}<f(x)<b_{j} \}= f − 1 ( I j ) = { x ∈ R : f ( x ) ∈ ( a j , b j )} = { x ∈ R : a j < f ( x ) < b j } = = { x ∈ R : f ( x ) > a j } ⏟ medible pues f medible ∩ { x ∈ R : f ( x ) < b j } =\underbrace{ \{ x \in \mathbb{R}:f(x)>a_{j} \} }_{ \text{medible pues f medible} }\cap \{ x \in \mathbb{R}:f(x)<b_{j} \} = medible pues f medible { x ∈ R : f ( x ) > a j } ∩ { x ∈ R : f ( x ) < b j } Luego, f − 1 ( I j ) f ^{-1}(I_{j}) f − 1 ( I j )
Ejercicio 3
Ejercicio 11 guía 9:
Queremos ver que si C = { x ∈ E : f ( x ) ≠ g ( x ) } ⟹ μ ( C ) = 0 \mathcal{C}=\{ x \in E:f(x)\neq g(x) \}\implies \mu(\mathcal{C})=0 C = { x ∈ E : f ( x ) = g ( x )} ⟹ μ ( C ) = 0 f f f g g g h = f − g h=f-g h = f − g
C = { x ∈ E : h ( x ) ≠ 0 } = { x ∈ R : h ( x ) > 0 } ⏟ C 1 ∪ { x ∈ R : h ( x ) < 0 } ⏟ C 2 \mathcal{C}=\{ x \in E:h(x)\neq 0 \}=\underbrace{ \{ x \in \mathbb{R}:h(x)>0 \} }_{ \mathcal{C_{1}} }\cup \underbrace{ \{ x \in \mathbb{R}:h(x)<0 \} }_{ \mathcal{C_{2}} } C = { x ∈ E : h ( x ) = 0 } = C 1 { x ∈ R : h ( x ) > 0 } ∪ C 2 { x ∈ R : h ( x ) < 0 } A su vez, C 1 = ⋃ n = 1 ∞ C 1 , n \mathcal{C_{1}}= \displaystyle \bigcup_{n=1}^{\infty}\mathcal{C_{1,n}} C 1 = n = 1 ⋃ ∞ C 1 , n C 1 , n = { x ∈ R : h ( x ) ≥ 1 n } \mathcal{C_{1,n}}=\left\{ x \in \mathbb{R}: h(x)\geq \frac{1}{n} \right\} C 1 , n = { x ∈ R : h ( x ) ≥ n 1 }
Como h h h ⟹ C 1 , n \implies\mathcal{C_{1,n}} ⟹ C 1 , n ∀ n ∈ N \forall n \in \mathbb{N} ∀ n ∈ N A = C 1 , n A=\mathcal{C_{1,n}} A = C 1 , n
∫ C 1 , n f = ∫ C 1 , n g ⟹ ∫ C 1 , n f − ∫ C 1 , n g = 0 ⟹ por linealidad ∫ C 1 , n h = 0 \int_{\mathcal{C_{1,n}}} f=\int_{\mathcal{C_{1,n}}}g\implies \int_{\mathcal{C_{1,n}}}f-\int_{\mathcal{C_{1,n}}}g=0\underset{ \text{por linealidad} }{ \implies }\int_{\mathcal{C_{1,n}}}h=0 ∫ C 1 , n f = ∫ C 1 , n g ⟹ ∫ C 1 , n f − ∫ C 1 , n g = 0 por linealidad ⟹ ∫ C 1 , n h = 0 Entonces
0 = ∫ C 1 , n h ( x ) d μ ( x ) ≥ m o n o t o n i a ∫ C 1 , n 1 n d μ ( x ) = ∫ E 1 n ⋅ X C 1 , n ⋅ d μ ( x ) = 1 n ⋅ μ ( C 1 , n ) ≥ 0 0=\int_{\mathcal{C_{1,n}}}h(x)\,d\mu(x)\underset{ monotonia }{ \geq }\int_{\mathcal{C_{1,n}}} \frac{1}{n}\: d\mu(x)=\int_{E} \frac{1}{n}\cdot\mathcal{X}_{\mathcal{C_{1,n}}}\cdot d\mu(x)= \frac{1}{n}\cdot \mu(\mathcal{C_{1,n}})\geq 0 0 = ∫ C 1 , n h ( x ) d μ ( x ) m o n o t o nia ≥ ∫ C 1 , n n 1 d μ ( x ) = ∫ E n 1 ⋅ X C 1 , n ⋅ d μ ( x ) = n 1 ⋅ μ ( C 1 , n ) ≥ 0 Luego μ ( C 1 , n ) = 0 ∀ n ∈ N \mu(\mathcal{C_{1,n}})=0\quad\forall n\in \mathbb{N} μ ( C 1 , n ) = 0 ∀ n ∈ N
0 ≤ μ ( C 1 ) = μ ( ⋃ n = 1 ∞ C 1 , n ) ≤ ∑ i = 1 ∞ μ ( C 1 , n ) = 0 0\leq \mu(\mathcal{C_{1}})=\mu\left( \bigcup_{n=1}^{\infty} \mathcal{C_{1,n}} \right)\leq \sum_{i=1}^{\infty} \mu(\mathcal{C_{1,n}})=0 0 ≤ μ ( C 1 ) = μ ( n = 1 ⋃ ∞ C 1 , n ) ≤ i = 1 ∑ ∞ μ ( C 1 , n ) = 0 Luego μ ( C 1 ) = 0 \mu(\mathcal{C_{1}})=0 μ ( C 1 ) = 0 C 2 = ⋃ C 2 , n \mathcal{C_{2}=\bigcup \mathcal{C_{2,n}}} C 2 = ⋃ C 2 , n C 2 , n = { x ∈ R : h ( x ) ≤ − 1 n } \mathcal{C_{2,n}}=\left\{ x \in \mathbb{R}:h(x)\leq - \frac{1}{n} \right\} C 2 , n = { x ∈ R : h ( x ) ≤ − n 1 } μ ( C 2 , n ) = 0 ∀ n ∈ N \mu(\mathcal{C_{2,n}})=0\quad\forall n\in \mathbb{N} μ ( C 2 , n ) = 0 ∀ n ∈ N μ ( C 2 ) = 0. \mu(\mathcal{C_{2}})=0. μ ( C 2 ) = 0.
μ ( C ) = μ ( C 1 ) + μ ( C 2 ) = 0 ⟹ f = g ctp en R \mu(\mathcal{C})=\mu(\mathcal{C_{1}})+\mu(\mathcal{C_{2}})=0\implies f=g\quad \text{ctp en } \mathbb{R} μ ( C ) = μ ( C 1 ) + μ ( C 2 ) = 0 ⟹ f = g ctp en R Citas y Comentarios
En el segundo parcial la guia 9 solo entra hasta el ejercicio 7 inclusive.