Convergence of a function involving a characteristic function of a decreasing interval Let $f_k: \mathbb{R} \rightarrow \mathbb{R} , f_k(x) = \frac{1}{\sqrt{x}}\chi_{\left[\frac{1}{2^{k+1}},\frac{1}{2^k}\right]}(x)$
For $k \rightarrow \infty $ the interval $[\frac{1}{2^{k+1}},\frac{1}{2^k}]$ will decrease it's range, but I would say that for every $k \in \mathbb{N}$ there exists an $x \in \mathbb{R}$ so that $x \in [\frac{1}{2^{k+1}},\frac{1}{2^k}]$.
Can anyone tell me the function $f \leftarrow f_k$ to which the sequence converges and how? Pointwise of course, even of course not. But in measure or in $L^1, L^2 ,...$?

You need to fix $x>0$. Then there is $K$ such that $\frac{1}{2^k}<x$, for $k>K$. Then $f_k(x)=0$, for $k>K$. Then $f(x)=0$.
For $x\leq0$, $f_k(x)=0$, for all $k$. Then $f(x)=0$.
Now, for $L^1$ we can integrate.
$$||f_k||_1=\int_{1/2^{k+1}}^{1/2^{k}}x^{-1/2}=\left.\frac{x^{1-1/2}}{1-1/2}\right|_{1/2^{k+1}}^{1/2^k}=2\left(\frac{1}{2^{k/2}}-\frac{1}{2^{(k+1)/2}}\right)\to0$$
This tells us that $||f_k-0||_1\to0$.
We can do the computation in general, in $L^p$:
$$||f_k||_p^p=\int_{1/2^{k+1}}^{1/2^{k}}x^{-p/2}=\begin{cases}\left.\frac{x^{1-p/2}}{1-p/2}\right|_{1/2^{k+1}}^{1/2^k}=\frac{1}{1-p/2}\left(\frac{1}{2^{k(1-p/2)}}-\frac{1}{2^{(k+1)(1-p/2)}}\right)&\text{ for }p\neq2\\\log(x)|_{1/2^{k+1}}^{1/2^k}=\log(1/2^k)-\log(1/2^{k+1})=\log(2)&\text{ for }p=2\end{cases}$$
The difference is now, that if $1-p/2\leq0$ then the limit doesn't go to zero. 
We know that for a convergent sequence in $L^p$ we can always extract a subsequence that converges pointwise a.e. Therefore if it were to converge it would converge to $f=0$. Therefore, for $1-p/2\leq0$ it doesn't converge. For $1-p/2>0$ it tends to $f=0$ as before.
