existence of Lebesgue integral

Hi, I have a question on my homework.

For each positive integer $n$, let $f_n:\mathbb{R}\to\mathbb{R}$ be integrable, $~f_n\geq 0$ and $f_n(x)\to f$ pointwise. I need to show that if $\int f_n$ converges to some finite $c\geq 0$, then $\int f$ exists and $0\leq\int f\leq c$.

I am thinking that for an arbitrary function $f$ , the Lebesgue integral exists iff $f$ is Lebesgue integrable or $\int f$ is infinite (is this correct?). However, for a nonnegative function $f$, the Lebesgue integral always exists and $\int f = \sup\{\int g：0\leq g\leq f, ~~g$ bounded and supported on a set of finite measure$\}$.

If what I am thinking is correct, then the question seems quite straigtforward. We have $f_n(x)$ converges to $f(x)$ for every $x$ and $f_n\geq 0$ for every $n$. So $f$ is nonnegative everywhere and $\int f$ exists since the Lebesgue integral exists for all nonnegative functions. That $0\leq\int f\leq c$ just follows from the fact that $f \leq 0$ and Fatou's Lemma.

Can someone tell me under what condition the Lebesgue integral exists? Is my attempt correct?

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I guess the downvotes come from poor formatting. Do you know how to put LaTeX? – Davide Giraudo May 26 '12 at 9:28
i havent learnt laytex... Can any one help me to edit it ? – keji May 26 '12 at 9:38
"...Lebesgue integral exist iff $f$ is Lebesgue integrable..." The definiiton of Lebesgue integrability is that the Lebesgue integral exists. – Austin Mohr May 26 '12 at 9:47
I edited it to make it look pretty in LaTeX (if the edit goes through). I'm also relatively new around here but the way I understand it it's considered a bit impolite to copy words like "Show that", since you're asking a question, not making a request. That might have played into the down vote as well. – Eric Stucky May 26 '12 at 9:58
Let $A \subset [0,1]$ be a set that is not Lebesgue measurable and let $1_A$ be its characteristic function. Then formally, $\int 1_A = \lambda(A)$ so that its integral does not exist. But it is contained in $[0,1]$, so it it not because is measure is infinity. – guaraqe May 26 '12 at 10:05

Depending on your course, showing that "the Lebesgue integral exists" could include showing that the integral (which can be defined for any non-negative measurable function) is finite. But it doesn't matter here: for any non-negative measurable (you didn't mention this; see also Juan's comment) function $g$, $\int g$ can be defined, and Fatou's lemma holds in that context, so that finiteness of the integral is established by afterwards by $$\int f\leq c.$$

In short: Your approach is correct.

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