# Proving for a Lebesgue measure the convergence in measure is equivalent to the convergence almost everywhere.

Hi I am working on the following question:

Let a sequence $\{f_k:[0,1]\to\mathbb{R}\}$ be a sequence of Lebesgue measurable functions. Assume that $\{f_k\}$ converges in measure to a function $g$ and simultaneously converges almost everywhere to a function $f$. Are $f$ and $g$ equivalent.

In this problem I am not quite sure what does $f$ and $g$ equivalent mean? I am thinking I have to prove that convergence in measure implies convergence pointwise and vice versa. If it is true I know how to prove convergence in measure implies convergence pointwise, but don't know the opposite direction.

Any help would be highly appreciated.

We need to show that $f=g$ a.e.

For this, it suffices to show that $|f - g| \leq 1/n$ a.e. for $n \in \mathbb{N}$ because then $\cup_n \{|f - g| > 1/n \}$ has measure $0$.

Fix $n_0 \in \mathbb{N}$ and suppose for contradiction that the set $A = \{|f-g| > 1/n_0 \}$ has positive measure.

Define a sequence of measurable subsets of $A$ by

$$A_N = \{x \in A: |f_n(x) - f(x)| \leq 1/2n_0, \forall n \geq N \}.$$

Now, $f_n \to f$ a.e. implies that almost every $x \in A$ belongs to some $A_N$. Therefore, $\cup_N A_N$ is equal to $A$ modulo a set of measure $0$, so $$\mu(\cup_N A_N) > 0,$$ since $A$ has positive measure.

By the continuity of measures, we must have $\mu(A_{N_0}) > 0$ for some $N_0$ (note that the sequence $(A_N)$ is increasing in $N$).

Now, by the triangle inequality $$1/n_0 < |f(x) - g(x)| \leq |f_n(x) - f(x)| + |f(x) - g(x)| \leq 1/2n_0 + |f(x) - g(x)|$$ for all $x \in A_{N_0}$ and all $n \geq N_0$.

But this contradicts the assumption that $f_n$ converges to $g$ in measure.