I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=X_{n}$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some quasi-independence property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)

I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=E(X_{n})$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some quasi-independence property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)

I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=X_{n}$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some sort of martingale property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)

I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=X_{n}$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some quasi-independence property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)

I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=X_{n-1}$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some sort of martingale property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)

I am a number theorist by profession, so apologies if the answer to this question is "trivially true" or "trivially false".

Let $(\Omega, \mathcal{A}, P)$ be a (non-atomic) probability space. Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration such $\mathcal{F}_\infty = \mathcal{A}$. Let $(X_n)_{n \geq 1}$ be a sequence of non-negative and bounded (but not uniformly bounded) random variables, such that $X_n$ is $\mathcal{F}_n$-measurable and $E(X_n | \mathcal{F}_{n-1})=X_{n}$. Let $B$ denote the set where $\sum_{n=1}^\infty X_n = \infty$.

Question: is it true that $P(B)$ can only be 0 or 1?

(To me, this seems to be a sort of Kolmogorov's zero-one law, where the independence is replaced by some sort of martingale property. However, I cannot see if the answer should be "yes" here, or if some additional integrability conditions or something of that sort would be necessary.)