Consider $$D=|D|=\frac{\sum_{i=1}^n \sum_{j=1}^n I[|{\sqrt{T}}\hat\theta_{ij,T}|>f(n)]}{g(n)}, $$ where $i \ne j$, $I$ is the indicator function, $ |{\hat\theta_{ij,T}}|<1$, $f(n)=O(\ln(n))$, $g(n)=O(n^2)$, and $f(n)/T=o(1)$.

The aim is to show that $D$ converges almost surely to $0$.

Taking the expectation, I know the following:

$$ E|D|=\frac{\sum_{i=1}^n \sum_{j=1}^n Pr[|{\sqrt{T}\hat\theta_{ij,T}}|>f(n)]}{g(n)}\, {\le}\,\,{\sup_{ij}}\,Pr[|{\sqrt{T}\hat\theta_{ij,T}}|>f(n)] \,\, {\le} \,\, {e^{-\frac{f(n)}{\rho}}} $$

where ${\rho}={\sup_{ij}}\, {\rho_{ij}}$ is positive and bounded.

From the Markov inequality: $$ Pr(|D|>{\epsilon})\,\,{\le}\,\,\frac{E|D|}{\epsilon} \,\,{\le}\,\, \frac{{e^{-\frac{f(n)}{\rho}}}}{\epsilon} $$ for some positive $\epsilon$ and hence $\lim_{n,T\to\infty} Pr(|D|>{\epsilon})=0.$

Is this result sufficient for $D$ to converge almost surely to 0? If not, how can I show that $D$ converges almost surely to 0?