Suppose $A$ is a $C^*$-algebra,an automorphism $\alpha$ on $A$ is called approximately inner if for every finite subset $F$ of $A$ and for every $\epsilon > 0$,there is an inner automorphism $\beta$ on $A$ such that $\|\alpha(a)-\beta(a)\|<\epsilon$ for all $a\in F$.
There is a proposition:If $A$ is separable and if $\alpha$ is an automorphism on $A$,then $\alpha$ is approximately inner iff there exits a sequence $\{\alpha_n\}$ in $Inn(A)$ such that $\alpha_n(a)\to \alpha(a)$ for all $a\in A$.
$\Longleftarrow$, for any finite set $F$ of $A$,since there exits a sequence $\{\alpha_n\}$ in $Inn(A)$ such that $\alpha_n(a)\to \alpha(a)$ for all $a\in A$.we can take $\alpha_{n_0}$ such that $\alpha_{n_0}(a)\to \alpha(a)$ for any $a\in F$
How to prove the right direction $\Longrightarrow$?
Hint: Let $\{a_n\}$ be a countable dense subset of $A$. For each $n\in\mathbb N$, there is an inner automorphism $\beta_n$ such that $$\|\alpha(a_k)-\beta_n(a_k)\|<\frac1n$$ for $1\leq k\leq n$.