Finite Measure Space Problem

Homework Statement

I have a sequence of functions converging pointwise a.e. on a finite measure space, $$\int_X |f_n|^p \leq M (1 < p \leq \infty$$ for all n. I need to conclude that $$f \in L^p$$ and $$f_n \rightarrow f$$ in $$L^t$$ for all $$1 \leq t < p.$$

The Attempt at a Solution

By Fatous I can show $$f \in L^p$$ and since $$L^t \subseteq L^p$$ for finite measure spaces, I have everything in L^t as well. I can apply Egoroffs to get $$\int_E |f_n-f|^t < \epsilon$$ with $$\mu(X-E) < \delta$$. Any ideas on how to proceed? And thanks for your time!

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Maybe try something like

$$\int_X{|f_n-f|^t}=\int_{X\setminus E}{|f_n-f|^t}+\int_{E}{|f_n-f|^t}$$

Try to find an upper bound K of $|f_n-f|^t$. Then the first integral becomes

$$\int_{X\setminus E}{|f_n-f|^t}\leq \mu(X\setminus E) K\leq \varepsilon K.$$

This is what I've been trying, but I cannot see a reason why |f_n-f|^t should be bounded on this set.

OK, here's how to proceed:

First, prove that

$$\lim_{\lambda \rightarrow 0}{\ \ \sup_n{\int_{\{|f_n|^t\geq \lambda\}}{|f_n|^t}}}=0$$

Hint: if $0<a<b$, then $b^t=b^{t-p}b^p\leq a^{t-p}b^p$

Second, prove that for each $\varepsilon >0$, there exists a $\delta>0$ such that for each E with $\mu(E)<\delta$, we have

$\int_E{|f_n|^t}<\varepsilon$

for each n.