If $\lim_{n\to\infty} \int_E f_n dx$ is finite, prove $f$ is Lebesgue integrable on $E$ and $\lim_{n\to\infty} \int_E f_n dx=\int_E fdx$.

Let $\{f_n\}_{n=1}^\infty$ be a decreasing sequence of Lebesgue integrable functions defined on a measurable set $E$. Suppose there is a function $f$ such that $f_n(x)\to f(x)$ almost everywhere on $E$.

If $\lim_{n\to\infty} \int_E f_n dx$ is finite, then prove $f$ is Lebesgue integrable on $E$ and $\lim_{n\to]infty} \int_E f_n dx=\int_E fdx$.

I know that since $E$ is measurable, $E^c$ is measurable. I thought that since $f_n\to f$, then $\exists k\in\mathbb{Z}_+$ such that if $n\geq k$ then $|f_n(x)-f(x)|<1/2$. Is this useable?

I'm having a hard time with this one. Any help would be greatly appreciated.

As $(f_n)$ is decreasing, this means $(f_1-f_n)$ is increasing and non-negative. Thus the monotone convergence theorem yields the claim.
• We don't know $|f_n|$ is dominated by a nonnegative integrable function, so I don't see how you apply DCT. – zhw. Apr 21 '15 at 18:40
The functions $f_1- f_n$ increase to $f_1-f.$ By the MCT, $$\int (f_1-f_n) \to \int(f_1-f).$$ But the integrals on the left have a finite limit. It follows that $f_1-f\in L^1,$ and we have $f_1-f_n \to f_1-f$ in $L^1.$ Therefore $f_n \to f$ in $L^1.$