= Solution
The <dominated convergence theorem> states that if measurable functions converge almost everywhere and their absolute values are bounded by a single <Lebesgue integrable function>, their limit is integrable and their integrals converge to its integral.
Put $b_j=a_j/j$. The hypothesis makes $b_j$ summable. On the <positive integers> with <counting measure> define $f_n(j)=(j/n)b_j\mathbf1_{j\leq n}$. For each fixed $j$ it tends to zero and $0\leq f_n(j)\leq b_j$. Dominated convergence gives
$$
\boxed{\frac1n\sum_{j=1}^n a_j=\sum_jf_n(j)\longrightarrow0,}
$$
the stated <Kronecker lemma>.
For a bounded interval union $B$, $S_j\sim N(0,j)$ has density bounded by $1/\sqrt{2\pi j}$. Hence $\mathbb P(S_j\in B)\leq |B|/\sqrt{2\pi j}$, and
$$
\mathbb E\sum_{j=1}^\infty\frac{\mathbf1_B(S_j)}j
\leq\frac{|B|}{\sqrt{2\pi}}\sum_{j=1}^\infty j^{-3/2}<\infty.
$$
By <Tonelli's theorem>, the nonnegative series is finite <almost surely>. <Kronecker lemma> then makes the empirical occupation fraction of $B$ tend to zero. The family of finite rational-endpoint interval unions is countable, so intersecting their probability-one events proves simultaneous convergence for all of them. For the whole line, every indicator is one. Thus
$$
\boxed{m(B)=0\text{ for bounded }B\in F_0,\qquad m(\mathbb R)=1,\quad\text{almost surely}.}
$$
There is no countably additive <Borel measure> extending these values: $\mathbb R=\bigcup_{n\geq1}(-n,n)$ would have measure zero by <continuity> from below, contrary to its assigned measure one. The escaped empirical mass cannot be captured by a <probability> measure on the real line.
This is the <Gaussian random walk spends zero density of time in bounded sets> phenomenon.
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