Consider the Lebesgue measure space $(\mathbb{R},\mathcal{M}(\mathbb{R}),m)$. Let us consider the sequence of functions $f_n:[0,1]\to \mathbb{R}$ defined by $f_n(x)=x^n$ for all $x\in [0,1]$ and for all $n\in \mathbb{N}$. Then $f_n$ converges pointwise $m$-a.e. on $[0,1]$ to $f\equiv 0$. I want to show that for all $E\subset [0,1]$ with $m(E)=0$, $f_n$ can not converge uniformly to $f$ on $[0,1]\setminus E$. Please help!

You basically wish to show that $f_n \not\to 0$ in $L^\infty[0,1]$.
For any $0 < \varepsilon < 1$ we have that $$\{x \in [0,1] : |f_n(x)| \ge \varepsilon\} = [\sqrt[n]{\varepsilon}, 1]$$
which has positive measure so $\|f_n\|_\infty \not\le \varepsilon$ for all $n \in \mathbb{N}$. In fact, $\|f_n\|_\infty = 1$.
More explicitly, assume that $f_n \to 0$ uniformly on $[0,1]\setminus E$. Then for $0 < \varepsilon < 1$ there exists $n \in \mathbb{N}$ such that $\sup_{x \in [0,1]\setminus E}|f_n(x)| < \varepsilon$. But $\{x \in [0,1] : |f_n(x)| \ge \varepsilon\} = [\sqrt[n]{\varepsilon}, 1]$ has positive measure so $\{x \in [0,1] : |f_n(x)| \ge \varepsilon\} \not\subseteq E$ which implies that $\sup_{x \in [0,1]\setminus E}|f_n(x)| \ge \varepsilon$.
For all $\delta\in (0,1)$, $\lambda\left((1-\delta,1)\cap \left([0,1]\setminus E\right)\right)$ is positive because $$\delta=\lambda\left(1-\delta,1\right)=\lambda\left((1-\delta,1)\cap \left([0,1]\setminus E\right)\right)+\underbrace{\lambda\left((1-\delta,1)\cap E\right)}_{=0}$$
Thus the set $[0,1]\setminus E$ necessarily contains a sequence which converges to $1$.