Did B[0,1] is a closed subspace of $L^2[0,1]$ Did $B[0,1]$={$f:[0,1]\Rightarrow \mathbb{R}$ is bounded } is a closed subspace of $L^2[0,1]$
I need to answer this in my Homework but How can it be that a sequence of bounded function is convegent to some function in $L^2$ which is not bounded? 
Any equivalence class in $L^2$ fill some function as large as we want in some subset of zero measure.
Do u think like me that there is a problem in the exercise?  
my answer: 
First , Let $f\in B\left(0,1\right)$so there exist some $M>0$such
that $\vert f\vert<M$ so $\int_{\left[0,1\right]}\vert f\vert^{2}d\lambda<M^{2}<\infty so$$f\in L_{2}\left(0,1\right)\Rightarrow B\left(0,1\right)\subset L_{2}\left(0,1\right)$
. 
Now,$\forall n\geq1$ define $f_{n}\left(x\right)=$$\begin{cases}
^{4}\sqrt{n} & x<\frac{1}{n}\\
0 & else
\end{cases}$ , $\vert f_{n}(x)\vert\leq n\Rightarrow f_{n}\in B\left(0,1\right)$.
$\left\{ f_{n}\right\} _{n=1}^{\infty}$is a Cauchy sequence .
Let $\epsilon>0$$m>n\geq1$:
$\Vert f_{n}-f_{m}\Vert_{2}=\int_{\left[0,1\right]}\vert f_{n}-f_{m}\vert^{2}d\lambda=\int_{\left[0,1\right]}\vert f_{n}-f_{m}\vert^{2}d\lambda<\int_{\left[0,\frac{1}{m}\right]}\vert n-m\vert^{2}d\lambda<\int_{\left[0,\frac{1}{m}\right]}\sqrt{m}d\lambda=1/\sqrt{m}
$
so $\forall n>m>\frac{1}{\lfloor\epsilon^{2}\rfloor}+1$ it takes
place that $\Vert f_{n}-f_{m}\Vert_{2}<\frac{1}{\sqrt{m}}<\epsilon$
but $f_n(0)\Rightarrow\infty$
 A: Presumably the set means the set of bounded measurable functions.
Here is a specific example:
Let $f_n(x) = {1 \over \sqrt[4]{x}} 1_{[{1 \over n},1]}(x)$ and $f(x) = {1 \over \sqrt[4]{x}} 1_{(0,1]}(x)$ and
$\|f-f_n\| = {4 \over n}$.
Each $f_n$ is bounded, but $f$ is not.
More generally, if $f \in L^2[0,1]$, then the function
$f_n(x) = \max(-n,\min(n, f(x))$ is bounded and we see that
$f_n \to f$ (in $L^2[0,1]$) by the dominated convergence theorem.
Hence $L^2[0,1]$ is the closure of the bounded (measurable) functions using the
$L^2[0,1]$ topology.
Note:
The notions of equivalence classes can sometimes cloud thinking of such things.
In this case, one has to ask what a bounded function means, since the equivalence class of a bounded function contains 
functions that are not bounded (this makes some assumptions about
the underlying measure, I am assuming the Lebesgue measure here).
Hence, strictly speaking, the set of bounded functions is not a subset of
$L^2[0,1]$, and what is meant is the set
$\{[f] | f \text{ is bounded} \}$, where $[f]$ denotes the equivalence
class.
