Cauchy sequence in $L^p$, existence of a set with finite measure, and integral is less than epsilon over the complement The setting is as follows, for each $f\in L^p$, define a set function $\lambda_f$ on $\mathcal{A}$ by $$\lambda_f(E)=\left(\int_E |f|^p d\mu\right)^{1/p}=\|f\cdot I_E\|_p.$$
Let $(f_n)$ be a Cauchy sequence in $L^p$. Show that for all $\epsilon >0$, there exists a set $E\in\mathcal{A}$ (depending on $\epsilon$) with $\mu (E)<\infty$ such that for all $F\in\mathcal{A}$ with $F\subseteq E^c$, we have $\lambda_{f_n} (F)<\epsilon$ for all $n\in\mathbb{N}$.
I tried proving by contradiction: Suppose to the contrary for all $E\in\mathcal{A}$ with $\mu(E)<\infty$, $\exists F\in\mathcal{A}$ with $F\subseteq E^c$ but $\lambda_{f_n}(F)\geq\epsilon$ for all $n\in\mathbb{N}$.
Intuitively, I see that that may be a problem as $\mu(E)$ gets larger, $\lambda_{f_n}(F)$ should tend towards zero? However I don't know how to prove it rigorously.
Thanks for any help.
Note: I set a bounty to attract more answers, so that I can choose one that I can understand. Having some problems with understanding this one. 
 A: As mentioned by Svetoslav, the key word here is "uniform integrability".
First observe that the result holds true if you take just one function $f$ instead of a sequence (or if you prefer, when your sequence $(f_n)$ is constant). 
This is not completely trivial if you do not want to use any theorem. For every $k\in\mathbb N$, set $E_k:=\{ 1/k\leq \vert f\vert\leq k\}$. The sequence $(E_k)$ is increasing, and the complement of $E_\infty:=\bigcup_k E_k$ is equal to $\{ f=0\}\cup\{ \vert f\vert=\infty\}$. Since $\vert f(x)\vert<\infty$ almost everywhere (because $f\in L^p$), it follows that $\int_{E_\infty^c} \vert f\vert^p=0$. Now, consider the finite measure $\mu_f$ defined by $\mu_f(A):=\int_A \vert f\vert^p$. Since $\mu_f$ is finite and $(E_k)$ is increasing, $\mu_f(E_k^c)\to \mu_f(E_\infty^c)=0$. So one can find $k$ such that $\mu_f(E_k^c)<\varepsilon^p$, i.e. $\lambda_f(E_k^c)<\varepsilon$. Finally, $E_k$ has finite $\mu$-measure because $\vert f\vert\geq 1/k$ on $E_k$ and $f\in L^p$ (formally, apply Markov's inequality to $\vert f\vert^p$). So you can take $E:=E_k$ for this suitably chosen $k$.
From the "one function case", it follows easily that one can find a suitable set $E$ for any finite family of functions $f_1,\dots ,f_N$.
Now, let us come back to our Cauchy sequence $(f_n)$, and assume that the result does not hold true for some $\varepsilon >0$. By the previous remark, this means that for any set $E$ with $\mu(E)<\infty$, one can find arbitrarily large $n\in\mathbb N$ such that $\lambda_{f_n}(E^c)\geq\varepsilon)$.
Since $(f_n)$ is Cauchy, one can find $N\in\mathbb N$ such that $\Vert f_n-f_N\Vert_p<\varepsilon/2$ for all $n\geq N$; then a set $E$ with $\mu(E)<\infty$ such that $\lambda_{f_N}(E^c)<\varepsilon/2$ (by the "one function case"); and then $n\geq N$ such that $\lambda_{f_n}(E^c)\geq\varepsilon$. But this is a contradiction since, by Minkowski's inequality, we have 
$$\lambda_{f_n}(E^c)\leq \lambda_{f_n-f_N}(E^c)+\lambda_{f_N}(E^c)\leq \Vert f_n-f_N\Vert_p+\lambda_{f_N}(E^c)<\varepsilon\, . $$
A: If $f \in L^p$ and $\epsilon>0,$ then there is a simple function $s, 0\le s \le |f|^p,$ such that $\int_X(|f|^p-s) < \epsilon.$ Now $s\in L^1,$ so no nonzero value of $s$ can be taken on a set of infinite measure. Because $s$ has only finitely many values, the set $E=\{s>0\}$ has finite measure. Thus
$$\int_{E^c}|f|^p = \int_{E^c}(|f|^p-s)+ \int_{E^c}s \le \int_{X}(|f|^p-s)+ 0 < \epsilon.$$
Now suppose $f_n$ is Cauchy in $L^p$ and $\epsilon >0.$ Then we can choose $N$ such that $\int_X|f_n-f_N|^p <\epsilon/2^p$ for $n>N.$ Because $|f_1|^p+ \cdots +|f_N|^p \in L^1,$ from the above we can find $E$ such that $\mu(E) <\infty$ and
$$\int_{E^c}(|f_1|^p+ \cdots +|f_N|^p) < \epsilon/2^p.$$
That handles $n=1,\dots , N.$ Suppose $n>N.$ Using $(a+b)^p \le 2^{p-1}(a^p+b^p)$ for $a,b\ge 0,$ we have
$$\int_{E^c}|f_n|^p \le 2^{p-1}(\int_{E^c}|f_n-f_N|^p +\int_{E^c}|f_N|^p) \le 2^{p-1}(\int_{X}|f_n-f_N|^p + \epsilon/2^p) < \epsilon.$$
Above I was assuming $1\le p < \infty,$ but I used no properties of $L^p$ as a normed linear space. The only place I used this restriction was the inequality $(a+b)^p \le 2^{p-1}(a^p+b^p).$ But for $0<p <1$ we have $(a+b)^p \le a^p+b^p.$ So actually the above works for the full range $0<p<\infty.$
