Let $\left\{[\frac{k}{2^{n}},\frac{k+1}{2^{n}}]:n\in\mathbb{N},k=0,...,2^{n}-1 \right\}=\left\{I_{n,k}:n\in\mathbb{N},k=0,...,2^{n}-1 \right\}$ and $f_{n}(x)=X_{I_{n,k}}$ can someone help to prove that $f_{n}(x)$ converges in measure to zero.
The definition that we gave is that , if $\forall \epsilon>0,\delta>0$ there exists $n_{0}$ such that $\forall n\geq n_{0}$ ,$\mu(\left\{x:\left | f_{n}(x)-f(x) \right |\geq \epsilon \right\})<\delta$
By the definition you gave of $f_n$ then the $x\in \mathbb{R}$ such that $| f_{n}(x)| > \epsilon $ are exactly the $x\in \mathbb{R}$ such that $f_n(x) = 1$, as $Im(f_n) = \{0,1\}$. When is $f_n(x) = 1$? This happens only when $x \in I_{n,k}$. And the measure of an interval is the difference of the end points. In this case the measure of $I_{n,k}$ is $\frac{1}{2^{n}}$. Note that it is a decreasing sequence that tends to $0$. Knowing this you can make it arbitrarily small if you choose a certain $n_0 \in \mathbb{N}$ such that $\frac{1}{2^{n_0}} < \epsilon$.