Let $f_n \in L_p(X, \mathbb{X}, \mu)$, $1 \leq p < \infty$, and let $\beta _n$ defined for $E \in \mathbb{X}$ by Let $f_n \in L_p(X, \mathbb{X}, \mu)$, $1 \leq p < \infty$, and let  $\beta _n$ defined for $E \in \mathbb{X}$ by
$\beta _n = (\int_E |f_n|^pd \mu)^{\frac{1}{p}}$ 
and suppose that $f_n$ is a Cauchy sequence. If $\epsilon >0$, then there exists a $\delta(\epsilon) >0$ such that if $E \in \mathbb{X}$ and $\mu(E) < \delta(\epsilon)$, then $\beta_n (E) < \epsilon$ for all $n \in \mathbb{N}$.
Hint: Use that if $f \in M^+$, and $\lambda$ is defined on $\mathbb{X}$ by  $\lambda (E) = \int_E fd \mu$, then  $\lambda$ is a measure and is absolutely continuous with respect to $\mu$. 
This is the 6.S exercise of Bartle the elements of integration and Lebesgue Measure. I thought it would only be an application from the previous exercise (6.R), but my teacher said that this exercise is very difficult, so I have no idea how to solve it. Someone can help
Note: I was able to solve the exercises below. Can these be applied in solving this?

 A: Fix $\varepsilon\gt 0$.
By 6. Q, there exists an integer $N$ such that for all $m,n\geqslant N$ and $E\in\mathbb X$;  $\left\lvert \beta_m\left(E\right)-\beta_n\left(E\right)\right\rvert\lt\varepsilon/2$. In particular, for all $n\geqslant N$, $\left\lvert \beta_n\left(E\right)-\beta_N\left(E\right)\right\rvert\lt\varepsilon/2$.
The hint suggest the following:

For any function $f\in\mathbb L^p$ and any $\varepsilon>0$, there exists $\delta=\delta\left(f,\varepsilon\right)$ such that if $E\in\mathbb X$ is such that  $\mu(E)\lt \delta$, then $\left\lvert f\chi_E\right\rVert_p\lt\varepsilon/2$. 

It suffices to treat the case $p=1$ and $f$ non-negative; to go to the general case, consider $\widetilde f:=\left\lvert f\right\rvert^p$. We then approximate $f$ in $\mathbb L^1$ by a bounded function.
To conclude, we let $\delta:=\min_{1\leqslant k\leqslant N}\delta\left(f_k,\varepsilon\right)$.
A: I came across this exercise in a measure theory and integration course and I actually solved it using only theorems, lemmas and corolaries from Bartle itself.
First, notice that the measures
$$\lambda_{n}(E) = \left\{ \int_{E} |f_{n}|^{p} d\mu \right\}^{1/p}, \ n \in \mathbb{N}$$
are absolutely continuous with respect to the measure $\mu$(this is a direct aplication of corolaries 4.9, 4.10 and 4.11).
Now, you have an equivalence of being absolutely continuous with respect to another measure $\mu$ given by:
Let $\lambda$ and $\mu$ be finite measures in a $\sigma$- algebra $\mathcal{A}$. Then $\lambda$ is absolutely continuous with respect to $\mu$ if and only if $\forall \varepsilon > 0$ there exists $\delta(\varepsilon) > 0$ such that $\forall X \in \mathcal{A}$ with $\mu(X) < \delta(\varepsilon)$ we have $\lambda(X) < \varepsilon$.
If you apply this equivalence you get that $\forall \varepsilon > 0$, $\exists X \in \mathcal{A}$ such that if $\mu(X) < \delta(\varepsilon)$ we have $\lambda_{n} < \varepsilon$, which translates as
$$\lambda_{n}(X) = \left\{ \int_{X}|f_{n}|^{p} \right\}^{1/p} < \varepsilon, \ \forall n \in \mathbb{N}$$
or,
$$\int_{X}|f_{n}|^{p} < \varepsilon^{p}, \ \forall n \in \mathbb{N}$$
as we wanted.
I strongly think that this equivalence is in Bartle but as I couldn't find it there so there you go!
To prove the equivalence mentioned, one of the sides is easy, let $\lambda$ satisy the condition, then if $\mu(X) = 0$ we have that $\lambda(X) < \varepsilon$ for every $\varepsilon > 0$, proving that $\lambda(X) = 0$ and so $\lambda$ is absolutely continuous with respct to $\mu$.
Now suppose that there exists $\varepsilon > 0$ and sets $X_{n} \in \mathcal{A}$ such that $\mu(X_{n}) < 2^{-n}$ and $\lambda(X_{n}) \geqslant \varepsilon$. Let $Y_{n} = \underset{k = n}{\overset{\infty}{\bigcup}}X_{k}$. That being said, notice that
$$\mu(Y_{n}) = \sum_{j = n}^{\infty} \mu(X_{j}) < \sum_{j = n}^{\infty}2^{-j} = 2^{-n} \sum_{l = 0}^{\infty}2^{-l} = 2^{-(n - 1)}$$
and $\lambda(Y_{n}) \geqslant \varepsilon$. As $(Y_{n})$ is a decreasing sequence of measurable sets we have that
$$\mu(\bigcap_{n = 1}^{\infty} Y_{n}) = \lim \mu(Y_{n}) = 0$$
and
$$\lambda(\bigcap_{n = 1}^{\infty} Y_{n}) = \lim \lambda(Y_{n}) \geqslant \varepsilon$$
that way, $\lambda$ is not absolutely continuous with respect to the measure $\mu$.
