A more abstract approach: Let $\theta$ be a shift-invariant probability mean on $\mathbb{N}$ (i.e., a finitely additive but not necessarily $\sigma$-additive set function with total weight 1 such that $\theta(\{n : n+1 \in A\}=\theta(A)$ for every subset $A$ of $\mathbb{N}$). (Such a mean can be obtained e.g. by taking a subsequential limit of the functions sending $A$ to $\frac{1}{n}\sum_{x\in[0,n]}1(x\in A)$.)

Let's define $A_+=\{n : a_1 + \cdots + a_n >0\}$, $A_0=\{n : a_1 + \cdots + a_n =0\}$ and $A_-=\{n : a_1 + \cdots + a_n <0\}$.

• The values of $\theta(A_+)$ and $\theta(A_-)$ are non-random by e.g. the Hewitt-Savage 0-1 law.
• By symmetry, $\theta(A_+)=\theta(A_-)$.
• $\theta(A_0)=0$ for every choice of $\theta$ since the upper density of $A_0$ is zero .

It follows that $\theta(A_+)=\frac{1}{2}$ almost surely.

(here's a reference from google for the third item above https://books.google.ca/books?id=eFFyBgAAQBAJ&lpg=PA16&ots=srDM7KjW76&dq=translation%20invariant%20means%20and%20upper%20density&pg=PA16#v=onepage&q=translation%20invariant%20means%20and%20upper%20density&f=false)

EDIT: As pointed out by Anthony Quas below, this approach suffers from what looks to be a quite serious measurability issue.

A more abstract approach: Let $\theta$ be a shift-invariant probability mean on $\mathbb{N}$ (i.e., a finitely additive but not necessarily $\sigma$-additive set function with total weight 1 such that $\theta(\{n : n+1 \in A\}=\theta(A)$ for every subset $A$ of $\mathbb{N}$). (Such a mean can be obtained e.g. by taking a subsequential limit of the functions sending $A$ to $\frac{1}{n}\sum_{x\in[0,n]}1(x\in A)$.)

Let's define $A_+=\{n : a_1 + \cdots + a_n >0\}$, $A_0=\{n : a_1 + \cdots + a_n =0\}$ and $A_-=\{n : a_1 + \cdots + a_n <0\}$.

• The values of $\theta(A_+)$ and $\theta(A_-)$ are non-random by e.g. the Hewitt-Savage 0-1 law.
• By symmetry, $\theta(A_+)=\theta(A_-)$.
• $\theta(A_0)=0$ for every choice of $\theta$ since the upper density of $A_0$ is zero .

It follows that $\theta(A_+)=\frac{1}{2}$ almost surely.

(here's a reference from google for the third item above https://books.google.ca/books?id=eFFyBgAAQBAJ&lpg=PA16&ots=srDM7KjW76&dq=translation%20invariant%20means%20and%20upper%20density&pg=PA16#v=onepage&q=translation%20invariant%20means%20and%20upper%20density&f=false)