Prove that $f_n \to 0$ in measure on $[0,1]$ $\iff$ $\lim_{n \to \infty}\int_{0}^{1}\mathbb{e}^{-|f_n(x)|^2} =1$ Prove that $f_n \to 0$ in measure on $[0,1]$ $\iff$ $\lim_{n \to \infty}\int_{0}^{1}\mathbb{e}^{-|f_n(x)|^2} =1$
My proof: 
Forward: Since $\mathbb{e}^{-x^2}$ is continuous we know that $\mathbb{e}^{-|f_n(x)|^2}\to 1$ in measure. Thus by DCT $\mathbb{e}^{-|f_n(x)|^2}\leq1$ we get the limit.
Backwards: Assume towards contradiction that $f_n \not \to 0$, then $\mu(x| |f_{n_k}|>\delta)>\epsilon$ for some $\delta,\epsilon$. Name these sets $A_k$ Thus $\int_{0}^{1}\mathbb{e}^{-|f_{n_k}(x)|^2}=\int_{A_k}\mathbb{e}^{-|f_{n_k}(x)|^2}+\int_{A_k^c}\mathbb{e}^{-|f_{n_k}(x)|^2}\leq\epsilon\mathbb{e}^{-\delta^2}+1-\epsilon<1$ contradiction
IS my proof correct? Is there a more direct way?

The definition of convergence in measure: For all $\delta>0$, 
$$
\lim_{n\to\infty} \mu \{x:\, |f_n(x)|>\delta\}=0.
$$
In other words, for all $\epsilon>0$, $\delta>0$, there exists $N=N(\epsilon,\delta)$ such that for all $n\geq N$,
$$
\mu \{x:\, |f_n(x)|>\delta\}<\epsilon.
$$
The negation is there exists $\epsilon_0>0$ and $\delta_0$ such that for all $k$, there exists $n_k\geq k$ such that 
$$
\mu \{x:\, |f_{n_k}(x)|>\delta_0\}\geq \epsilon_0.
$$
Hence, I believe your backward direction is correct with some recommendations. You should make a remark that since the limit exists, so does
$$
\lim_{k\to\infty}\int_0^1 e^{-|f_{n_k}(x)|^2}\,dx\leq \epsilon e^{-\delta^2}+1-\epsilon<1.
$$
As you have it right now, it looks like you are just showing that some terms are bounded by something less than 1. I know what you mean, but I would be more explicit with the limits.
