Two questions about convergence in measure I am currently studying for my analysis comprehensive exam and have a few questions about convergence in measure.
First of all I know that for a sequence $\{f_n\} \in L^p(E)$, $1 \le p < \infty$, $\{f_n\}$ converges to $f$ in $L^p(E)$ implies convergence in measure by using Chebyshev's inequality.
I am less sure about the $p = \infty$ case.  Here is what I have so far.  Since $\{f_n\}$ converges to $f$ in $L^\infty(E)$  after possibly excising a set of measure zero we may assume that $\{f_n\}$ converges uniformly to $f$ on E.  Then if $E$ has finite measure this would say that $\{f_n\}$ converges to $f$ in measure.  Is this argument valid?  Also I am not sure what to do in the case where E is not of finite measure, I suspect that it may end up being false but have not yet found a counterexample.
Secondly I have proven the following: If $\{f_n\}$ converges to $0$ in measure on $E$ of finite measure then $\lim \int_E \frac{|f_n|}{1 + |f_n|} = 0$
However my proof utilizes the Lebesgue dominated convergence theorem with the assumption of point wise a.e. convergence replaced with convergence in measure.  I know this is valid since it is an exercise in my book but I am not sure how to prove it.  Since there is a strong likely hood that a problem like this might show up on my exam it would be nice if I could get either a quick proof of the statement that avoids Lebesgue dominated convergence or a quick proof that convergence in measure is enough for Lebesgue dominated that I could reproduce on my exam.
Thanks in advance for any help you can offer.
 A: Convergence in $L^\infty$ does imply convergence in measure, even if $E$ has infinite measure.
To see this, suppose that $\{f_n\}$ converges to $f$ in $L^\infty$. Let $\epsilon > 0$. Then there exists an $N_\epsilon$ such that $\|f_n - f\|_\infty < \epsilon$ for all $n > N_\epsilon$.
Therefore, for all $n > N_\epsilon$, we see that $|f_n - f| < \epsilon$ almost everywhere, so $\mu\{x \in E : |f_n(x) - f(x)| > \epsilon\} = 0$. This is true for all $n > N_\epsilon$, which means that
$$\lim_{n \rightarrow \infty} \mu\{x \in E : |f_n(x) - f(x)| > \epsilon\} = 0$$
which means that $f_n \rightarrow f$ in measure.

Regarding the second question, put $g_n = |f_n| / (1 + |f_n|)$. Choose any $\epsilon > 0$. Then there is an $N_\epsilon$ such that for all $n > N_\epsilon$ we have $\mu(E_n) < \epsilon$, where
$$E_n = \{x \in E : |f_n(x)| > \epsilon\}$$
Let $n > N_\epsilon$ and define
$$t = \frac{\epsilon}{1 + \epsilon}$$
Note that $g_n > t$ if and only if $|f_n| > \epsilon$, and $t \rightarrow 0$ if and only if $\epsilon \rightarrow 0$. Then
$$\int_E g_n = \int_{E_n} g_n + \int_{E \setminus E_n} g_n$$
Since $g_n \leq 1$, and $\mu(E_n) < \epsilon$, the first integral is bounded above by $\epsilon$.
Since $g_n < t$ in $E \setminus E_n$, the second integral is bounded above by $t\mu(E\setminus E_n) \leq t\mu(E)$. The finiteness of $\mu(E)$ means that $t \rightarrow 0$ implies $t\mu(E) \rightarrow 0$, and the conclusion follows:
$$\lim_{n \rightarrow \infty} \int_E g_n = 0$$
