Uniqueness of limiting functions Let $\{f_n\},\ f_n:[a,b]\mapsto\mathbb{R}$. Suppose that the limiting function, $f(x)$, of the sequence satisfies
$$
\lim\limits_{n\rightarrow\infty}\int_a^b|f_n(x)-f(x)|^2dx = 0.
$$
Is this limiting function unique?
 A: The existence of the integral seems to indicate that $f_{n}(x)$ and $f(x)$
are square integrable. The $f_{n}(x)$ need not be continuous, they can be
simple functions
\begin{equation*}
f_{n}(x)=\sum_{k}c_{nk}\chi _{M_{k}}(x),\;M_{k}\cap M_{l}=0,\;k\neq l,\;\cup
M_{k}=[a,b]
\end{equation*}
where $\chi _{M_{k}}(x)$ is the characteristic function for the (measurable)
set $M_{k}$. Any $L^{2}$-function is the limit in $L^{2}$-norm of such a
sequence in the almost everywhere (ae) sense. In this sense $f(x)$ is unique
provided it exists which seems to be assumed. In general it need not have
any continuity properties. Note that this result is somewhat more general
than the statement by  user1952009 who assumed continuity or boundedness of
the $f_{n}$'s.
A: If we have that $\lim_{n\to\infty}\int_a^b|f_n-g|dx=0$, we get that $\int_{a}^{b}|f-g|dx\le \lim_{n\to\infty}\int_a^b|f-f_n|dx+\lim_{n\to\infty}\int_a^b|f_n-g|dx=0$. So $|f-g|$ is a positive function of vanishing integral. This means that if $f, g$, and thus $|f-g|$ are continuous, then $f=g$. But in general $f$ and $g$ can differ on a set of measure zero. For for example, let $f_n=0$, and $f=0$ and $g(x)=0$ if $x\notin\mathbb{Q}$, $g(x)=x$ else, Then since $\int_a^b |g(x)|dx=0$, we have that $f_n\to f$ and $f_n\to g$.
