Properties of a pseudo-metric on a measure space Given a measure space $\left(X,\mathcal{F},\mu\right)$
  and two $\mathcal{F}-\mbox{measurable}$
  functions $f,g:\left(X,\mathcal{F}\right)\to\mathbb{R}$
  we define the following:$$d\left(f,g\right)=\inf_{a>0}\left(a+\mu\left(\left\{ x\in X\;|\;\left|f\left(x\right)-g\left(x\right)\right|>a\right\} \right)\right)$$
 It can be shown that $d$
  is symmetric and admits the triangle inequality. I'm trying to show that given two functions $f,g:\left(X,\mathcal{F}\right)\to\mathbb{R}$
  and $\alpha\in\mathbb{R}$
  : $$d\left(\alpha f,\alpha g\right)\leq\max\left\{ 1,\left|\alpha\right|\right\} \cdot d\left(f,g\right)$$
 I've tried going at it from a couple of directions but I always get stuck with an inequality involving $\mu\left(\left\{ x\in X\;|\;\max\left\{ 1,\left|\alpha\right|\right\} \left|f\left(x\right)-g\left(x\right)\right|>a\right\} \right)$
  and no way to extract that pesky maximum outside. 
Also regardless of this question I think that convergence relative to $d$ is equivalent to convergence in measure relative to $\mu$ is that true?
Help would be appreciated.
 A: Let us answer

Also regardless of this question I think that convergence relative to $d$ is equivalent to convergence in measure relative to $\mu$ is that true?

first. Yes, that is true.
Let $f_n \to f$ in measure. Let $\varepsilon > 0$ be arbitrary. By the convergence in measure, there is an $N$ such that $\mu(\{ x: \lvert f(x) - f_n(x)\rvert > \varepsilon/2\}) < \varepsilon/2$ for all $n \geqslant N$. But then we have
$$d(f,f_n) \leqslant \frac{\varepsilon}{2} + \mu\left(\left\lbrace x : \lvert f(x) - f_n(x)\rvert > \frac{\varepsilon}{2}\right\rbrace\right) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$
for $n \geqslant N$. So convergence in measure implies $d$-convergence. Conversely, let $d(f,f_n)\to 0$, and let $\varepsilon > 0$ be arbitrary. We need to show that $\mu(\{ x : \lvert f(x) - f_n(x)\rvert \geqslant \varepsilon\}) \to 0$. Let $\delta > 0$ be arbitrary, subject to the restriction $\delta < \varepsilon$. Since $d(f,f_n) \to 0$, there is an $N$ with $d(f,f_n) < \delta/2$ for $n \geqslant N$. By the definition of $\inf$, there is then for each $n \geqslant N$ an $a_n > 0$ with
$$a_n + \mu(\{ x : \lvert f(x) - f_n(x)\rvert > a_n\}) < \delta.$$
But then we have $a_n < \delta$, and $\mu(\{ x : \lvert f(x) - f_n(x)\rvert > a_n\}) < \delta$, and since $a_n < \delta < \varepsilon$, that implies $\mu(\{ x : \lvert f(x) - f_n(x)\rvert \geqslant \varepsilon\}) < \delta$. That holds for all $n \geqslant N$, hence $f_n \to f$ in measure.
Regarding the inequality
$$d(\alpha f,\alpha g) \leqslant \max \{1,\lvert\alpha\rvert\}\cdot d(f,g),$$
the case $\alpha = 0$ is clear ($d(0,0) = 0$), and since multiplication by $-1$ leaves the absolute modulus unchanged, the case for negative $\alpha$ follows from that for positive $\alpha$.
Let first $0 < \alpha \leqslant 1$. Then
$$\begin{align}
d(\alpha f,\alpha g) &= \inf_{a > 0} \left(a + \mu(\{ x : \lvert \alpha f(x) - \alpha g(x)\rvert > a\}) \right)\\
&= \inf_{a>0} \left(a + \mu(\{ x : \lvert f(x) - g(x)\rvert > a/\alpha\right)\\
&= \inf_{b > 0} \left(\alpha b + \mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\
&\leqslant  \inf_{b > 0} \left(b + \mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\
&= d(f,g).
\end{align}$$
For $\alpha > 1$, the computation is almost the same, only in the end we don't eliminate $\alpha$ and instead multiply the measure with $\alpha$,
$$\begin{align}
d(\alpha f, \alpha g) &= \dotsb\\
&= \inf_{b > 0} \left(\alpha b + \mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\
&\leqslant  \inf_{b > 0} \left(\alpha b + \alpha\mu(\{x : \lvert f(x) - g(x)\rvert > b\})\right)\\
&= \alpha d(f,g).
\end{align}$$
A: Right I took another crack at it and I think I might have proven the claim in my second question. I'm not confident about the proof so I'd appreciate it if someone could review it.
From the definition of $d\left(f_{n},f\right)$
  we know that for all $\varepsilon>0$
  : $$d\left(f_{n},f\right)\leq\varepsilon+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>\varepsilon\right\} \right)$$
 Since $f_{n}\to f$
  in measure we know that for all $\varepsilon>0$
  there is an $N$
  such that for all $n>N$
 :$$\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>\varepsilon\right\} \right)<\varepsilon$$
 So for all $n>N$
  we can see that $d\left(f_{n},f\right)<2\varepsilon$
  and thus $d\left(f_{n},f\right)\overset{n\to\infty}{\longrightarrow}0$
 .
On the other hand given $\varepsilon>0$
  from the definition of $d\left(f_{n},f\right)$
  for each $n\in\mathbb{N}$
  there is an $a\left(n\right)>0$
  such that:$$a\left(n\right)+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)\leq d\left(f_{n},f\right)+\frac{\varepsilon}{2}$$
 Since $d\left(f_{n},f\right)\overset{n\to\infty}{\longrightarrow}0$
  there is an $N$
  such that $d\left(f_{n},f\right)<\frac{\varepsilon}{2}$
  for all $n>N$
  and thus for all $n>N$:$$a\left(n\right)+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)\leq d\left(f_{n},f\right)+\frac{\varepsilon}{2}<\varepsilon
 $$ Since this is true for all $\varepsilon>0$
  we get that:$$\lim_{n\to\infty}\left(a\left(n\right)+\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)\right)=0$$
 Since this is a sum of two non-negative sequences we know that $a\left(n\right)\overset{n\to\infty}{\longrightarrow}0$
  and: $$\left(\star\right)\;\lim_{n\to\infty}\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(n\right)\right\} \right)=0$$
 Now given $a>0$
  there is an $N$
  such that $a\left(N\right)\leq a$
  and thus:$$\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{N}\left(x\right)-f\left(x\right)\right|>a\right\} \right)\leq\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{N}\left(x\right)-f\left(x\right)\right|>a\left(N\right)\right\} \right)$$
 Also for all $\varepsilon>0$
  from $\left(\star\right)$
  there is an $N^{'}\geq N$
  such that for all $n>N^{'}$
  we get:$$\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\right\} \right)\leq\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\left(N\right)\right\} \right)<\varepsilon$$
 Since this is true for all $\varepsilon>0$
  we get that for all $a>0$
 :$$\lim_{n\to\infty}\mu\left(\left\{ x\in\mathbb{R}\;|\;\left|f_{n}\left(x\right)-f\left(x\right)\right|>a\right\} \right)=0$$
 And thus $f_{n}\to f$
  in measure.
