Does $\lim_{n\rightarrow \infty} \int_X f_n - \int_X f\gt 0$ implies that convergence of $f_n$ to $f$ a.e. fails?

I've come across this problem as a part of another proof that I'm writing and I want to know if this is a right conclusion:

Let $X$ be a finite measure space and $\{f_n\}$ be a sequence of nonnegative integrable functions.

If I know: $$\lim_{n\rightarrow \infty} \int_X f_n - \int_X f \geq \delta > 0,$$ can I conclude that $\mu\{x: f_n \nrightarrow f\} > 0$ or in other words $f_n$ doesn't convege to $f\ a.e.$?

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If $X=[0,1]$ and $f_n=n\chi_{(0,n^{—1}})$, then $f_n\to 0$ almost everywhere, $f_n\geq 0$ but $\int_X f_n=1$. –  Davide Giraudo Oct 28 '12 at 21:59
And you have to assume $\delta>0$ (and the measure $>0$) in order to make the assertion more interesting :) –  Davide Giraudo Oct 28 '12 at 22:10

I assume you want to conclude that $\mu(\{x: f_n \not\to f\}) > 0$. The answer is NO.
Consider $$f_n = \begin{cases} n & x \in [0,1/n)\\ 0 & \text{otherwise} \end{cases}$$ and $$f = 0$$