# Question about the definition of convergence in measure.

In my text book convergence in measure ${{f}_{k}}\mathop \to \limits^{m} {f}$ is defined as "$\forall \epsilon> 0$ we have $\mathop {\lim }\limits_{k \to \infty } |\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > \epsilon \} | = 0$".

My problem is what happens to the case of "$\epsilon = 0$".

It looks to me that

$\mathop {\lim }\limits_{k \to \infty } |\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > 0\} | = 0$ can imply ${{f}_{k}}\mathop \to \limits^{m} {f}$ since $\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > \varepsilon \} \subseteq \{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > 0\}$ for any $k$.

For the other direction, since $\mathop {\lim }\limits_{k \to \infty } |\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > \varepsilon \} | = 0$ holds for any $\epsilon > 0$, isn't it

$\mathop {\lim }\limits_{k \to \infty } |\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > 0\} | = \mathop {\lim }\limits_{\varepsilon \to 0} \mathop {\lim }\limits_{k \to \infty } |\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > \varepsilon \} | = \mathop {\lim}\limits_{\varepsilon \to 0} 0 = 0$

Then why we don't define convergence in measure as $\mathop {\lim }\limits_{k \to \infty } |\{ x \in \Omega :\left| {{f_k}\left( x \right) - f\left( x \right)} \right| > 0 \} | = 0$?

Thank you!

• The first equality in the second long equation, you are interchanging two limits. This is not always true. – henryforever14 Aug 1 '15 at 16:23

## 1 Answer

Just look at the sequence $f_n(x) := \frac{1}{n}$ on the space $\Omega := [0,1]$ (endowed with Lebesgue measure). Since $f_n$ converges pointwise (even uniformly) to $f := 0$, we have in particular $f_n \to 0$ in measure. However, $$|\{x \in \Omega; |f_n(x)-f(x)|>0\}|=1,$$ so we cannot that this term converges to $0$ as $n \to \infty$.