1. Borel-Cantelli Lemma

You want to show that

$$P(\limsup A_n) = 0$$

where $A_n = \{|f_n| > n\} = \{\frac{|f_n|}{n} > 1\}$

Perhaps we might try Borel-Cantelli Lemma?

If we can show the following

$$\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} P(\frac{|f_n|}{n} > 1) < \infty$$

Note that $P(A_n^C) = 1 - P(\frac{|f_n|}{n} \le 1) $

then we would get what we want.

2. Kolmogorov 0-1 Law

Denote

$$\overline{f_n} := \frac{1}{n}\sum\limits_{i=1}^n f_i$$

to have:

$$P(\lim \overline{f_n}=c) >0$$

Observe that

$$\{\lim \overline{f_n}=c \} \in \bigcap_{n=1}^{\infty} \sigma(f_n, f_{n+1}, ...)$$

By the Kolmogorov 0-1 Law,

$$P(\lim \overline{f_n}=c) = 0 \ \text{or} \ 1$$

Since

$$P(\lim \overline{f_n}=c) >0$$

we have that

$$P(\lim [\overline{f_n}]=c) = 1$$

which is equivalent to either:

$$P(\liminf [\overline{f_n}]=c) = 1$$

$$P(\limsup [\overline{f_n}]=c) = 1$$

Try using one of those with one of the inequalities below.

3. Important inequalities (Probability w/ Martingales):

1, 2

3. $$\liminf x_n > z \to x_n > z \ \text{eventually}$$

4. $$\liminf x_n < z \to x_n < z \ \text{infinitely often}$$

1. Borel-Cantelli Lemma

You want to show that

$$P(\limsup A_n) = 0$$

where $A_n = \{|f_n| > n\} = \{\frac{|f_n|}{n} > 1\}$

Perhaps we might try Borel-Cantelli Lemma?

If we can show the following

$$\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} P(\frac{|f_n|}{n} > 1) < \infty$$

Note that $P(A_n^C) = 1 - P(\frac{|f_n|}{n} \le 1) $

then we would get what we want.

2. Kolmogorov 0-1 Law

Denote

$$\overline{f_n} := \frac{1}{n}\sum\limits_{i=1}^n f_i$$

to have:

$$P(\lim \overline{f_n}=c) >0$$

Observe that

$$\{\lim \overline{f_n}=c \} \in \bigcap_{n=1}^{\infty} \sigma(f_n, f_{n+1}, ...)$$

By the Kolmogorov 0-1 Law,

$$P(\lim \overline{f_n}=c) = 0 \ \text{or} \ 1$$

Since

$$P(\lim \overline{f_n}=c) >0$$

we have that

$$P(\lim [\overline{f_n}]=c) = 1$$

which is equivalent to either:

$$P(\liminf [\overline{f_n}]=c) = 1$$

$$P(\limsup [\overline{f_n}]=c) = 1$$

Try using one of those with one of the inequalities below.

3. Important inequalities

Williams - Probability with Martingales

Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

1. Borel-Cantelli Lemma

You want to show that

$$P(\limsup A_n) = 0$$

where $A_n = \{|f_n| > n\} = \{\frac{|f_n|}{n} > 1\}$

Perhaps we might try Borel-Cantelli Lemma?

If we can show the following

$$\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} P(\frac{|f_n|}{n} > 1) < \infty$$

Note that $P(A_n^C) = 1 - P(\frac{|f_n|}{n} \le 1) $

then we would get what we want.

2. Kolmogorov 0-1 Law

Denote

$$\overline{f_n} := \frac{1}{n}\sum\limits_{i=1}^n f_i$$

to have:

$$P(\lim \overline{f_n}=c) >0$$

Observe that

$$\{\lim \overline{f_n}=c \} \in \bigcap_{n=1}^{\infty} \sigma(f_n, f_{n+1}, ...)$$

By the Kolmogorov 0-1 Law,

$$P(\lim \overline{f_n}=c) = 0 \ \text{or} \ 1$$

Since

$$P(\lim \overline{f_n}=c) >0$$

we have that

$$P(\lim [\overline{f_n}]=c) = 1$$

which is equivalent to either:

$$P(\liminf [\overline{f_n}]=c) = 1$$

$$P(\limsup [\overline{f_n}]=c) = 1$$

Try using one of those with one of the inequalities below.

3. Important inequalities (Probability w/ Martingales):

1, 2

3. $$\liminf x_n > z \to \liminf(x_n > z)$$

4. $$\liminf x_n < z \to \limsup(x_n < z)$$

1. Borel-Cantelli Lemma

You want to show that

$$P(\limsup A_n) = 0$$

where $A_n = \{|f_n| > n\} = \{\frac{|f_n|}{n} > 1\}$

Perhaps we might try Borel-Cantelli Lemma?

If we can show the following

$$\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} P(\frac{|f_n|}{n} > 1) < \infty$$

Note that $P(A_n^C) = 1 - P(\frac{|f_n|}{n} \le 1) $

then we would get what we want.

2. Kolmogorov 0-1 Law

Denote

$$\overline{f_n} := \frac{1}{n}\sum\limits_{i=1}^n f_i$$

to have:

$$P(\lim \overline{f_n}=c) >0$$

Observe that

$$\{\lim \overline{f_n}=c \} \in \bigcap_{n=1}^{\infty} \sigma(f_n, f_{n+1}, ...)$$

By the Kolmogorov 0-1 Law,

$$P(\lim \overline{f_n}=c) = 0 \ \text{or} \ 1$$

Since

$$P(\lim \overline{f_n}=c) >0$$

we have that

$$P(\lim [\overline{f_n}]=c) = 1$$

which is equivalent to either:

$$P(\liminf [\overline{f_n}]=c) = 1$$

$$P(\limsup [\overline{f_n}]=c) = 1$$

Try using one of those with one of the inequalities below.

3. Important inequalities (Probability w/ Martingales):

1, 2

3. $$\liminf x_n > z \to x_n > z \ \text{eventually}$$

4. $$\liminf x_n < z \to x_n < z \ \text{infinitely often}$$