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Suppose that a sequence of integrable function $(f_n)$ converges uniformly to the zero function. I want to find an example of such $f_n$ exhibiting the property that $$\int f_n \not \to 0.$$

What if we replace uniform convergence with uniformly bounded and point wise convergence?

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    You should specify the limits of integration. If you integrate over a compact interval then you won't find such example. If you integrate over an unbounded one then there are examples. – 2017-02-17
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    @NeedForHelp Thanks, but that still doesn't help me too much – 2017-02-18
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    Have a look [here](http://math.stackexchange.com/questions/24171/lebesgue-integral-uniform-convergence) for a start. – 2017-02-18

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Does $f_n=\frac{1}{n}\chi_{[0,n]}$ work? The integral of all these are $1$, but they converge uniformly to the $0$ function.

If you want a compact domain (or even just a finite measure domain), and your functions $f_i$ are uniformly bounded and pointwise convergent, then you use Dominated Convergence theorem on the constant function that bounds them all.

Also, I don't know why your bounty answer needs to draw from a credible or official source. If you believe and understand the answer, why does it matter where it comes from?

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    What if I want a finite measure domain, point wise convergence of $f_n \to 0$, but not necessarily uniformly bounded? – 2017-02-19
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    Counterexample: take a bump function on $[0, 1]$, and then dilate it by $\frac{1}{2}$ in the $x$ direction and $2$ in the $y$ direction, and repeat. – 2017-02-19
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    @user3359 On $(0,1),$ let $f_n(x) = nx^n.$ Then $f_n\to 0$ pointwise on $(0,1),$ yet $\int_0^1 f_n = n/(n+1) \to 1.$ – 2017-02-19
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    @BobJones Can you give me an example of where the convergence is point wise, not uniform, but the functions are uniformly bounded? I'll then be happy to award you the bounty. – 2017-02-22
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    Try a bump function on $[0, 1]$ and then translate it to the right by $1$ repeatedly. – 2017-02-23
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    @BobJones What will the integral of the bump function converge to? – 2017-02-26
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    Whatever the integral of the original bump function is. It's constant. – 2017-02-26
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    @user3359 Why did I not receive the full bounty? Did you forget to accept, or were my answers not to your satisfaction? – 2017-02-27