$f_n$ converges to $f$ in measure iff $f_n-f$ converges to $0$ in measure.

Question

Prove that $f_n$ converges to $f$ in measure iff $f_n-f$ converges to $0$ in measure.

What I know:

$(f_n)$ is said to converge in measure to $f$ if:

$$m\{|f_n-f|\geq \epsilon\}<\epsilon \text{ for all } n\geq N$$

where $m(\cdot)$ denotes the Lebesgue measure.

I'm not sure where to start for the forward direction.

For the backwards direction:

Suppose $f_n-f$ converges to $0$ in measure. Then $m\{(f_n-f)-0\geq \epsilon\}<\epsilon$, which just simplifies to $m\{f_n-f\geq \epsilon\}<\epsilon$.

I'm new to the concept of convergence in measure so a hint would be preferred over a solution. Thanks.

Answer 1

Suppose that $f_n$ converges to $f$ in measure, i.e., for every $\epsilon >0$

\begin{equation}\lim_\limits{n\to \infty}\mu\{|f_n-f|>\epsilon\}=0. \end{equation} In particular, \begin{equation}\lim_\limits{n\to \infty}\mu\{|(f_n-f)-0|>\epsilon\}=0, \end{equation} which says that $f_n-f$ converges to $0$ in measure.