bounded variation functions on $[0,1]$ are always $L_2[0,1]$?

My question is $BV[0,1] \subset L_2[0,1]$ or not.

My own answer (not sure correct): $f \in BV[0,1]$ implies that removing discontinuity, we have continuous function $\tilde{f}$ such that \begin{align*} BV(f) = const + \int_{[0,1]} |\tilde{f}(x)|dx < \infty, \end{align*} where the constant is from discontinuity. As discontinuities have Lesbesgue measure zero, it is sufficient to show $\tilde{f}$ is square integrable.

But since $L_2[0,1] \subset L_1[0,1]$, we can find some $\tilde{f}$ that is NOT square-integrable.

Is it right?

BV functions are bounded. Since they can be written as the difference of two monotone functions, they are Borel measurable. Hence, they belong to $L^{p}$ for every $p >0$.