You can use the following result to modify the step function appropriately:
Suppose $f:[a,b] \to \mathbb{R}$ is integrable, with $|f(x)| \le M$ for some $M$, and $f$ is continuous on some open (relative to $[a,b]$) interval $I \subset [a,b]$, except at some point $c \in I$.
Then, for any $\epsilon>0$ and any $y$ such that $|y|\le M$, we can find some function $\tilde{f}$ such that (i) $\tilde{f}(x) = f(x)$ for $x \notin I$,
(ii) $\tilde{f}$ is continuous on $I$,
(iii) $|\tilde{f}(x)| \le M$
(iv) $\|f-\tilde{f}\| < \epsilon$ and
(v) $\tilde{f}(c) = y$.
If $c \in (a,b)$, choose $x_1,x_2 \in I$ such that $x_1 < c < x_2$ and
$|x_2-x_1| < ({1 \over 2M} \epsilon)^2$. Let
$\tilde{f}(x) = \begin{cases}
f(x_1)+ {x-x_1 \over c - x_1} (y-f(x_1)), & x \in [x_1,c] \\
c+ {x-c \over x_2 - c} (f(x_2)-c), & x \in [c,x_2] \\
f(x), & \text{otherwise}\end{cases}$. Then $\tilde{f}$ is continuous on $I$,
equal to $f$ everywhere else, bounded by $M$ and
$\|f-\tilde{f}\| = \sqrt{\int_{[x_1,x_2]}|f(x)-\tilde{f}(x)|^2 dx} < \epsilon$.
If $c=a$ or $c=b$, it should be clear how to modify the above accordingly.