If $Y_n to Y_infty$ a.s. and $P(Y_infty in D_f circ g)=0$ then $f(g(Y_n)) to f(g(Y_infty))$ a.s.

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I have encountered the following line in a proof from Durrett's Probability Theory and Examples. Here, we have $f$ continuous and $P(Y_infty in D_g)=0$ $D_g = x: g ;textis discontinuous at ; x$.

If $Y_n to Y_infty$ a.s. and $P(Y_infty in D_f circ g)=0$ then $f(g(Y_n)) to f(g(Y_infty))$ a.s.

How does this implication hold?

analysis probability-theory measure-theory convergence

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asked Jul 17 at 12:03

takecare

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up vote
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I have encountered the following line in a proof from Durrett's Probability Theory and Examples. Here, we have $f$ continuous and $P(Y_infty in D_g)=0$ $D_g = x: g ;textis discontinuous at ; x$.

If $Y_n to Y_infty$ a.s. and $P(Y_infty in D_f circ g)=0$ then $f(g(Y_n)) to f(g(Y_infty))$ a.s.

How does this implication hold?

analysis probability-theory measure-theory convergence

share|cite|improve this question

asked Jul 17 at 12:03

takecare

2,25811431

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I have encountered the following line in a proof from Durrett's Probability Theory and Examples. Here, we have $f$ continuous and $P(Y_infty in D_g)=0$ $D_g = x: g ;textis discontinuous at ; x$.

If $Y_n to Y_infty$ a.s. and $P(Y_infty in D_f circ g)=0$ then $f(g(Y_n)) to f(g(Y_infty))$ a.s.

How does this implication hold?

analysis probability-theory measure-theory convergence

share|cite|improve this question

asked Jul 17 at 12:03

takecare

2,25811431

I have encountered the following line in a proof from Durrett's Probability Theory and Examples. Here, we have $f$ continuous and $P(Y_infty in D_g)=0$ $D_g = x: g ;textis discontinuous at ; x$.

If $Y_n to Y_infty$ a.s. and $P(Y_infty in D_f circ g)=0$ then $f(g(Y_n)) to f(g(Y_infty))$ a.s.

How does this implication hold?

analysis probability-theory measure-theory convergence

share|cite|improve this question

asked Jul 17 at 12:03

takecare

2,25811431

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 12:03

takecare

2,25811431

asked Jul 17 at 12:03

takecare

2,25811431

asked Jul 17 at 12:03

takecare

2,25811431

2,25811431

add a comment | 

add a comment | 

1 Answer
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accepted

If $Y_n to Y_infty$ a.s., then $Y_n(omega) to Y_infty(omega)$ for all $omega$ in a set $F$ with $P(F)=1$. Consider the set $G = F cap (Y_infty in D_f circ g)^c$. If $omega in G$, then $Y_n(omega) to Y_infty(omega)$ and $f circ g$ is continuous at $Y_infty(omega)$, hence $f(g(Y_n(omega))) to f(g(Y_infty(omega)))$. Finally, note that $P(G)=1$.

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answered Jul 17 at 12:23

aduh

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

If $Y_n to Y_infty$ a.s., then $Y_n(omega) to Y_infty(omega)$ for all $omega$ in a set $F$ with $P(F)=1$. Consider the set $G = F cap (Y_infty in D_f circ g)^c$. If $omega in G$, then $Y_n(omega) to Y_infty(omega)$ and $f circ g$ is continuous at $Y_infty(omega)$, hence $f(g(Y_n(omega))) to f(g(Y_infty(omega)))$. Finally, note that $P(G)=1$.

share|cite|improve this answer

answered Jul 17 at 12:23

aduh

4,25031338

add a comment | 

up vote
1
down vote



accepted

If $Y_n to Y_infty$ a.s., then $Y_n(omega) to Y_infty(omega)$ for all $omega$ in a set $F$ with $P(F)=1$. Consider the set $G = F cap (Y_infty in D_f circ g)^c$. If $omega in G$, then $Y_n(omega) to Y_infty(omega)$ and $f circ g$ is continuous at $Y_infty(omega)$, hence $f(g(Y_n(omega))) to f(g(Y_infty(omega)))$. Finally, note that $P(G)=1$.

share|cite|improve this answer

answered Jul 17 at 12:23

aduh

4,25031338

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

If $Y_n to Y_infty$ a.s., then $Y_n(omega) to Y_infty(omega)$ for all $omega$ in a set $F$ with $P(F)=1$. Consider the set $G = F cap (Y_infty in D_f circ g)^c$. If $omega in G$, then $Y_n(omega) to Y_infty(omega)$ and $f circ g$ is continuous at $Y_infty(omega)$, hence $f(g(Y_n(omega))) to f(g(Y_infty(omega)))$. Finally, note that $P(G)=1$.

share|cite|improve this answer

answered Jul 17 at 12:23

aduh

4,25031338

If $Y_n to Y_infty$ a.s., then $Y_n(omega) to Y_infty(omega)$ for all $omega$ in a set $F$ with $P(F)=1$. Consider the set $G = F cap (Y_infty in D_f circ g)^c$. If $omega in G$, then $Y_n(omega) to Y_infty(omega)$ and $f circ g$ is continuous at $Y_infty(omega)$, hence $f(g(Y_n(omega))) to f(g(Y_infty(omega)))$. Finally, note that $P(G)=1$.

share|cite|improve this answer

answered Jul 17 at 12:23

aduh

4,25031338

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 12:23

aduh

4,25031338

answered Jul 17 at 12:23

aduh

4,25031338

answered Jul 17 at 12:23

aduh

4,25031338

4,25031338

add a comment | 

add a comment | 

 

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