exists $f\not\in L^1(X)$ such that $f\in L^p(X)$

Question

Definition:

Let $ (X,M,\mu)$ be a measure space. $ (X,M,\mu)$ is a $\sigma $-finite measure space iff $\exists\displaystyle A_{1},A_{2},\ldots \in M : \forall n\in \mathbb{N}: {\displaystyle \mu \left(A_{n}\right)<\infty }$ and ${\displaystyle \bigcup _{n\in \mathbb {N} }A_{n}=X}$

Problem

a) If $X$ is $\sigma$-finite, then $\exists{\displaystyle B_{1},B_{2},\ldots \in {M}} :\forall {\displaystyle n\in \mathbb {N} }:$ $1\leq {\displaystyle \mu \left(B_{n}\right)<\infty }$ and ${\displaystyle \bigcup _{n\in \mathbb {N} }B_{n}=X}$ and for ${\displaystyle i\neq j}$ ${\displaystyle B_{i}\cap B_{j}=\varnothing }$

b) Show that exists $f\not\in L^1(X)$ such that for all $1<p\leq\infty$, $f\in L^p(X)$

(As a help give us that: Take $f$ to be constant on each $B_k$)

Proof:

a) For part a) it is sufficient to use the following result:

if $D_1 \subseteq D_2 \subseteq ...$ and are in $M$ then $\mu (\bigcup_k D_k)$=lim $\mu(D_k)$, this to prove that:

-) $\exists k_1\in \mathbb {N}:$ if we define $B_1 :=\bigcup_{i=1}^{k_1} A_i$ then $\mu (B_1)\geq 1$.

-) $\exists k_2\in \mathbb {N}:$ if we define $B_2 :=(\bigcup_{i=k_1 +1}^{k_2} A_i)\setminus (B_1)$ then $\mu (B_2)\geq 1$.

-) $\exists k_3\in \mathbb {N}:$ if we define $B_3 :=(\bigcup_{i=k_2 +1}^{k_3} A_i)\setminus (B_1 \bigcup B_2)$ then $\mu (B_3)\geq 1$.

and so on with $B_n$, then, they fulfill part a).

My question

My question is regarding part b) of the problem, I have not yet been able to see how to use the help and part a) to define $f$, maybe it is not complicated, although I still can't do it, if someone has an idea, I would be grateful to share it.

UPDATE

consider: $f=\sum \frac{1}{n\mu(B_n)}\chi_{B_{n}}$

Note that: $1-p<0$ and as $1\leq {\displaystyle \mu \left(B_{n}\right)<\infty }$ then $[\mu(B_n)]^{1-p}\leq 1$

$\int \left [ \sum \frac{1}{n\mu(B_n)}\chi_{B_{n}} \right ]^p d\mu=\int \sum \left [ \frac{1}{n\mu(B_n)} \right ]^p \chi_{B_{n}} d\mu = \sum \left [\frac{1}{n\mu(B_n)} \right ]^p \int \chi_{B_{n}} d\mu= \sum \left [\frac{1}{n\mu(B_n)} \right ]^p \mu(B_n)= \sum \left [\frac{1}{n} \right ]^p [\mu(B_n)]^{1-p}\leq \sum \left [\frac{1}{n} \right ]^p$

for $p=1$ the sum diverges, then $f\not\in L^1(X)$

for $p>1$ the sum converges, then $f\in L^p(X)$

for $p=\infty$, $\left | f \right | \leq 1$, then $f\in L^{\infty}(X)$