On $||f||_μ := \inf {\{r : μ({\{x : |f(x)| > r }\}) ≤ r}\}$. The following is an exercise from Bruckner's Real Analysis:

For (c) I think : Because $\lim_{n \to \infty} μ({\{x : |f(x)-f_n(x)| > r }\}) = 0$ for any $r \ge 0$ thus the inf attains $0$; on the other direction, since inf is zero then the limit is zero?
For (e) I think it intuitively but I don't know how make it rigorous : if $c \ge 1$ then $||cf||_μ=c||f||_μ$ and for $c \le 1$ then $||cf||_μ=||f||_μ$.
I have no idea how to approach this problem especially the definition of $||f||_μ$ seems to be complicated enough to get relations based on that.
added- I add my attempt for parts (g) and (h):
Since $\sum_1^{\infty}||g_{k+1}-g_k||_{\mu}< \infty,$ then there is some $N \in \mathbb{N}$ such that for all $n>N$, and all $\epsilon>0$ : $\sum_1^{\infty}||g_{k+1}-g_k||_{\mu}< \epsilon $. By (b), $||g_n-g_m||_{\mu} \le \sum_{i=m}^{n-1}||g_{i+1}-g_i||_{\mu} < \epsilon$ thus ${\{g_n}\}_n$ is Cauchy and if $d(f,g)=||f-g||_{\mu}$ can be proved to be a metric which by previous parts only remains to prove that if $||f||_{\mu}=0$ then $f=0$ a.e. which I am not sure if it is true. But considering that it's true, there is some $g$ such that $||g_n-g||_{\mu} \to 0$. We construct that $g$ : Our ﬁrst step is to ﬁnd a certain subsequence. Given $ε = 2^{−(j+1)}$, there exists $n_j$ such that if $n, m \ge n_j$, then $||g_n −g_m||_{\mu} \le 2^{−(j+1)}$. WLOG we may assume $n_j \ge n_{j−1}$ for each $j$. Considering $n_0=0$, we hope the limit function will be $\sum_m (g_{n_m} − g_{n_{m−1}}).$  Then (using part (2)), $||g_j||_{\mu} \le \sum_{m=1}^j ||g_{n_m} − g_{n_{m−1}}||_{\mu} \le ||g_{n_1}||+ \frac12$. Hence $g$ is ﬁnite a.e. This proves the absolute convergence for almost every $x$. Set $g(x) = \sum_{m=1}^{\infty} (g_{n_m} − g_{n_{m−1}}) = \lim_K g_{n_K}$ since we have a telescoping series. But I couldn't figure out how to do for convergence in measure, though.
My question: Am I correct with parts (g) and (h)? Thanks.
 A: Well, this is not one question, but eight related questions. Here is the answer for them.
Let us go step by step. First, some notation: we will write $[|f|>r]$ to mean $\{ x : |f(x)|>r \}$.
Let $S= \{ r: \mu([|f|>r]) \leq r \}$. Then, since $\mu$ is non-negative, if $r\in S$ then $r \geq 0$. Note also that, for all $f$, $+\infty \in S$.
Item (a) Let $\{r_n\}_n$ be a non-increasing sequence of elements in $\{ r: \mu([|f|>r]) \leq r \}$ such that $r_n \searrow \|f\|_\mu$.
We have that, for all $n$, $\mu([|f|>r_n]) \leq r_n$. We also have that $[|f|>r_n] \nearrow [|f|>\|f\|_\mu]$. So
$$ \mu([|f|>\|f\|_\mu])= \lim_n \mu([|f|>r_n])\leq \lim_n r_n = \|f\|_\mu $$
So $ \mu([|f|>\|f\|_\mu]) \leq \|f\|_\mu $.
Item (b) Note that for any $\alpha, \beta >0$,
$$ [|f+g|> \alpha + \beta ] \subseteq [|f|> \alpha ] \cup [|g|> \beta ] $$
So, if $\alpha \in  \{ r: \mu([|f|>r]) \leq r \}$ and $\beta \in \{ r: \mu([|g|>r]) \leq r \} $, then
$$\mu( [|f+g|> \alpha + \beta ] ) \leq \mu( [|f|> \alpha ] ) + \mu( [|g|> \beta ] ) \leq \alpha + \beta$$
So, $\alpha+\beta \in \{ r: \mu([|f + g |>r]) \leq r \}$.
So, for all  $\alpha \in  \{ r: \mu([|f|>r]) \leq r \}$ and $\beta \in \{ r: \mu([|g|>r]) \leq r \} $, we have
$$ \|f+g\|_\mu  =\inf \{ r: \mu([|f + g|>r]) \leq r \} \leq \alpha + \beta $$
Let $\alpha_n \searrow   \inf \{ r: \mu([|f|>r]) \leq r \}  = \|f\|_\mu$ and
$\beta_n \searrow   \inf \{ r: \mu([|g|>r]) \leq r \}  = \|g\|_\mu$. For all $n$, we have
$$ \|f+g\|_\mu   \leq \alpha_n + \beta_n $$
Taking the limit as $n \to \infty$, we have $\|f+g\|_\mu \leq \|f\|_\mu + \|g\|_\mu $.
item (c) Note that
$f_n \to f$ in measure if and only if, for all $\varepsilon, \delta >0$, there is $N$ such that, if $n > N$, then $\mu([|f_n-f|> \varepsilon])\leq \delta$.
$\|f_n -f\|_\mu \to 0$ if and only if, for all $\varepsilon >0$, there is $N$ such that, if $n > N$, then $\mu([|f_n-f|> \varepsilon])\leq \varepsilon$.
It is immediate that $f_n \to f$ in measure $\Rightarrow $ $\|f_n -f\|_\mu \to 0$. Just take $\delta = \varepsilon$.
Now let us prove that  $\|f_n -f\|_\mu \to 0$ $\Rightarrow $ $f_n \to f$ in measure.
Suppose $\|f_n -f\|_\mu \to 0$. Then, for all $\varepsilon >0$, there is $N_\varepsilon$ such that, if $n > N_\varepsilon$, then $\mu([|f_n-f|> \varepsilon])\leq \varepsilon$.
Given any $\varepsilon, \delta >0$:

*

*If $\varepsilon < \delta$ we have that there is $N_\varepsilon$ such that, if $n> N_\varepsilon$, $\mu([|f_n-f|>\varepsilon])\leq \varepsilon < \delta$.


*If $\delta \leq \varepsilon$ we have that there is $N_\delta$ such that, if $n> N_\delta$, $\mu([|f_n-f|>\varepsilon])\leq \mu([|f_n-f|>\delta]) \leq \delta$.
So, for all $\varepsilon, \delta >0$, there is $N$ such that, if $n> N$, $\mu([|f_n-f|>\varepsilon]) \leq \delta$. So $f_n \to f$ in measure.
item (d):  Let $S= \{ r: \mu([|c\chi_A|>r]) \leq r \}$.
First, note that $\|0\|_\mu =0$. In fact, if $f =0$, then $\mu([|f|> 0]) = \mu(\emptyset) =0 \leq 0$. So, $0 \in S$. So $\|f\|_\mu=0$ (remember that all elements of $S$ are non-negative).
If $c=0$, then $c\chi_A=0$ and $\|c\chi_A\|_\mu = 0 = \inf\{ c, \mu(A)\}$.
If $c>0$, then for all $r \geq 0$

*

*if $r<c$ then $[|c\chi_A|>r] = A$ and then $ r \in S$ if and only if $\m(A) \leq r$;

*if $r \geq c$  then $[|c\chi_A|>r] = \emptyset$ and $r \in S$.

So, if $\mu(A) < c$ then $\|c\chi_A\|_\mu = \mu(A)$, and if $\mu(A) \geq  c$ then $\|c\chi_A\|_\mu = c$. So we have that $\|c\chi_A\|_\mu  = \inf\{ c, \mu(A)\}$.
item (e) Let $S_1=  \{ r: \mu([|f|>r]) \leq r \}$ and $S_c =  \{ r: \mu([|cf|>r]) \leq r \}$.
If $c < 1$ , then $[|cf|>r] \subseteq [|f|>r]$, so if $r \in S_1$, then
$\mu([|cf|>r]) \leq  \mu([|f|>r]) \leq r $
So $S_1 \subseteq S_c$. So
$$ \|c f \|_\mu = \inf S_c \leq \inf S_1 = \| f \|_\mu$$
If $c \geq 1$ , then $[|cf|>cr] = [|f|>r]$, so if $r \in S_1$, then
$\mu([|cf|>cr]) =  \mu([|f|>r]) \leq r \leq cr$, so $cr \in S_c$. So we have $cS_1 \subseteq S_1$ and so, we have
$$ \|c f \|_\mu = \inf S_c \leq \inf cS_1 = c\inf S_1= c\| f \|_\mu$$
So $\|c f \|_\mu  \leq \max \{ \| f \|_\mu, c\| f \|_\mu  \}$.
item (f) The statement of this item is missing one condition. As stated, it is false, with a trivial counter-example: let $A$ be a set such that $\mu(A) < \infty$, let $f=\infty \chi_A$. Then for any $c>0$, $cf =f$ and  then $\|cf \|_\mu$ does not converge to zero as $c \to 0$.
The condition missing is that $f$ must be finite a.e.. Suppose $f$ is finite a.e..
So let $B = [|f|=\infty]$. Since $f$ is finite a.e., we have $\mu(B)=0$.
Let $n \in \Bbb N$, we have $ [|f|>n] \subseteq [|f|\geq 0]$ and $ [|f|>n] \searrow B$. So, using the fact that $\mu([|f|\geq 0]) < \infty$, we have
$$ \mu([|f|>n]) \searrow \mu(B)=0$$
Now, given any $\varepsilon >0$, there is $N$ such that if $n>N$ then  $ \mu([|f|>n]) < \varepsilon$.
Take $\delta = \frac{\varepsilon}{N+1} $ . Then, for any $c$ such that $|c| < \delta$, we have
$$ [|cf|>\varepsilon]= [|f|>\varepsilon/|c|] \subseteq [|f|>\varepsilon/\delta]= [|f|>N+1]$$
So
$$ \mu( [|cf|>\varepsilon]) \leq \mu ([|f|>N+1]) \leq \varepsilon $$
So $\varepsilon \in \{ r: \mu([|cf|>r]) \leq r \}$. So $\|cf\|_\mu \leq \varepsilon$.
So we have prove that, for all  $\varepsilon >0$, there is $\delta >0$ such that, if $|c|<\delta$ then $\|cf\|_\mu \leq \varepsilon$.  That means $\|cf\|_\mu \to 0$ as $c \to 0$.
item (g)  Let us understand item (g).  It is rather trivial to adapt our solution to item (c) to show that $g_n$ is Cauchy in $\mu$-measure   if and only $g_n$ is Cauchy regarding $\|\cdot\|_\mu$. So, if $g_n$ is Cauchy regarding $\|\cdot\|_\mu$ then

*

*there is a function $g$ such that there is a sub-sequence $g_{n_k}$ converging to $g$ a.e. and $g_n$ converges to $g$ in  $\mu$-measure.

The "challenge" in this item is that if $\sum_{k=1}^\infty \|g_{k+1} - g_k\|_\mu  < \infty$ then, we have a condition stronger than being a Cauchy sequence  regarding $\|\cdot\|_\mu$ such that  we don't need to take the sub-sequence for the convergence a.e.. Here is a detailed proof:
Let $a_n = \|g_{k+1} - g_k\|_\mu $. From item (a), we have that
$$ \mu([|g_{k+1}-g_k|> a_k ]))\leq a_k$$
Now, note that , for all $r \leq k \leq s $, we have
$$\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \subseteq  \left [|g_s -g_k|> \sum_{i=k}^{s-1} a_i \right]\subseteq  \bigcup_{i=k}^{s-1} [|g_{i+1}-g_i|> a_i ] \subseteq \bigcup_{i=r}^\infty [|g_{i+1}-g_i|> a_i ]$$
So, for all $s,k$ such that $r \leq k \leq s$, we have
$$\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \subseteq  \bigcup_{i=r}^\infty [|g_{i+1}-g_i|> a_i ]$$
So, we have
$$\bigcup_{k \geq r} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \subseteq  \bigcup_{i=r}^\infty [|g_{i+1}-g_i|> a_i ]$$
So,
$$\mu \left (\bigcup_{k \geq r} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \right ) \leq  \mu \left (\bigcup_{i=r}^\infty [|g_{i+1}-g_i|> a_i ] \right ) \leq \sum_{i=r}^\infty a_i $$
So, for all $r\in \Bbb N$, we have
$$\mu \left (\bigcap_{t \in \Bbb N} \bigcup_{k \geq t} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \right ) \leq \mu \left (\bigcup_{k \geq r} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \right ) \leq   \sum_{i=r}^\infty a_i $$
Since $ \sum_{i=1}^\infty a_i < \infty$, we have that, for all $r\in \Bbb N$
$$\mu \left (\bigcap_{t \in \Bbb N} \bigcup_{k \geq t} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \right )= 0$$
So,
$$\mu \left (\bigcup_{r \in \Bbb N}\bigcap_{t \in \Bbb N} \bigcup_{k \geq t} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ] \right )= 0$$
Let $A = \bigcup_{r \in \Bbb N}\bigcap_{t \in \Bbb N} \bigcup_{k \geq t} \bigcup_{s \geq k}\left [|g_s -g_k|> \sum_{i=r}^\infty a_i \right ]$.
Now, note that,  if $x\in X$ and $x\notin A$, then, for each $r$,
there is $T \in \Bbb N$ such that, for all $s, k$, $T \leq k \leq s $, we have  $|g_s(x) -g_k(x)| \leq \sum_{i=r}^\infty a_i   $.
It follows that $g_n$ is Cauchy a.e., so there is a function $f$ such that $g_n$ converges to $f$ a.e..
However, from 1 in the beginning of this item, we know that there is a function $g$ such that there is a sub-sequence $g_{n_k}$ converging to $g$ a.e. and $g_n$ converges to $g$ in  $\mu$-measure. So $f=g$ a.e.. This concludes the proof of item (g). $\square$
Item (h) As stated this item does not mention $\| \cdot \|_\mu$ and it is a trivial consequence of theorems regarding convergence in measure and convergence a.e. (Bruckner 4.2).
I believe the intent in the item was to show that if $g_n$ is Cauchy in  $\| \cdot \|_\mu$, then there is a sub-sequence $g_{n_k}$ that converges both in measure and $\mu$-almost everywhere.
The easiest way to solve it is to adapt our solution of item (c) to show that $g_n$ is Cauchy in $\mu$-measure   if and only $g_n$ is Cauchy regarding $\|\cdot\|_\mu$, and then apply the theorems regarding convergence in measure and convergence a.e. (Bruckner 4.2).
Remark: Let $S= \{ r: \mu([|f|>r]) \leq r \}$. We know that, if $r\in S$, then $r \geq 0$. We also know that $+\infty \in S$.
Note that, given $r, t \in [0, +\infty]$, if $r \in S$ and $r \leq t$, we have
$ [|f|>t] \subseteq [|f|>r]$, so we have
$$ \mu([|f|>t] \leq \mu([|f|>r] \leq r$$
So  $t\in S$.
Thus, we have that $S$ is of the form $(a, +\infty]$ or $[a, +\infty]$, for some $a \geq 0$. So, by item (a), $S$ is always of the form $[a, +\infty]$, for some $a \geq 0$. Of course, $a$ is precisely $\|f\|_\mu$.
