I want to know whether the Borel Cantelli Lemma is true for a random walk. More precisely,this question can be described as follows.

Let $X_1,X_2,\cdots$ be i.i.d. taking values in $\mathbb{R}^d$ and let $S_n=X_1+\cdots+X_n$. Suppose that $A\subset\mathbb{R}^d$ is a Borel set (for example, $A$ can be $\{0\}$ or $(-\infty,0]$ for $d=1$), then do we have $P(S_n\in A,i.o.)=0$ if and only if $\sum\limits_{n=1}^\infty P(S_n\in A)<\infty$ ?
Where $i.o.$ means infinitely often, that is $(S_n\in A,i.o.)=\mathop{\cap}\limits_{k=1}^\infty\mathop{\cup}\limits_{n=k}^\infty(S_n\in A)$.

Obviously,the "$\Leftarrow$" part of this question is true because of the usual Borel-Cantelli Lemma.
In a special case, when $A=\{0\}$, this proposition is true, and we can see a proof from Rick Durrett's book Probability:Theory and Examples, Fourth Edition, Chapter 4.2. It is proved by introducing a series of stopping time: let $\tau_n$ be the time of the $n$th return to $0$, that is define $\tau_0=0$ and $\tau_n=\inf\{m>\tau_{n-1}:S_m=0\}$. But this method does not work for other cases of $A$.
I know that this proposition is true when :
(1) $A$ is a finite set ; (2) $A$ is a bounded set and 0 is an interior point of $A$.
For case (1) , we can imitate the method described above when $A=\{0\}$.
For case (2) , we have a theorem (also from Probability:Theory and Examples, Fourth Edition, Chapter 4.2 ):

If $\sum\limits_{n=1}^\infty P(\Vert S_n\Vert<\varepsilon)=\infty$, then $P(\Vert S_n\Vert<2\varepsilon, i.o.)=1$, and the convergence or divergence of the sums is independent of $\varepsilon$. ( where $\Vert\cdot\Vert$ is a norm on $\mathbb{R}^d$ )

But what about other cases?

I want to know whether the Borel-Cantelli lemma is true for a random walk. More precisely, this question can be described as follows.

Let $X_1,X_2,\cdots$ be i.i.d. taking values in $\mathbb{R}^d$ and let $S_n=X_1+\cdots+X_n$. Suppose that $A\subset\mathbb{R}^d$ is a Borel set (for example, $A$ can be $\{0\}$ or $(-\infty,0]$ for $d=1$), then do we have $P(S_n\in A,i.o.)=0$ if and only if $\sum\limits_{n=1}^\infty P(S_n\in A)<\infty$ ?
Where $i.o.$ means infinitely often, that is $(S_n\in A,i.o.)=\mathop{\cap}\limits_{k=1}^\infty\mathop{\cup}\limits_{n=k}^\infty(S_n\in A)$.

Obviously, the "$\Leftarrow$" part of this question is true because of the usual Borel-Cantelli Lemma.
In a special case, when $A=\{0\}$, this proposition is true, and we can see a proof from Rick Durrett's book Probability:Theory and Examples, Fourth Edition, Chapter 4.2. It is proved by introducing a series of stopping time: let $\tau_n$ be the time of the $n$th return to $0$, that is define $\tau_0=0$ and $\tau_n=\inf\{m>\tau_{n-1}:S_m=0\}$. But this method does not work for other cases of $A$.
I know that this proposition is true when :
(1) $A$ is a finite set ; (2) $A$ is a bounded set and 0 is an interior point of $A$.
For case (1) , we can imitate the method described above when $A=\{0\}$.
For case (2) , we have a theorem (also from Probability:Theory and Examples, Fourth Edition, Chapter 4.2 ):

If $\sum\limits_{n=1}^\infty P(\Vert S_n\Vert<\varepsilon)=\infty$, then $P(\Vert S_n\Vert<2\varepsilon, i.o.)=1$, and the convergence or divergence of the sums is independent of $\varepsilon$. ( where $\Vert\cdot\Vert$ is a norm on $\mathbb{R}^d$ )

But what about other cases?