Concerning the closedness of a subset of $L^1[0,1]$ I was reading some stuff about that involved $L^p$ spaces and came up with this question: Given $X=L^1[0,1]$, I need to show a closed, convex subset of a Banach space with some properties, the thing is that apparently the the set 
$$\mathcal{C}=\left\{f\in X:\int_0^1fdx=1\right\}$$
 is closed, and I just can't make sense of it, so there may be something I'm missing. Convexity is not a problem at all but the closed part.
For instance, take the sequence 
$$f_n=\left\{\begin{array}{cc}n&x\in[0,1/n]\\0&x\in[1/n,1]\end{array}\right. $$
 so $\displaystyle\int f_ndx=1$ for all $n$, but I'm kind of sure that there is no $f\in X$ such that $$||f_n-f||_1\to0$$ as $n\to\infty$ so I'm sort of stuck here. Any hints will be very well appreciated. Thanks a lot.

If $f_n \to f$ and $\int f_n= 1$ then
$|\int f - 1| = | \int (f-f_n)| \le \int |f-f_n| = \|f-f_n\|$. Hence
$\int f = 1$.
