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Find a sequence of functions $f_n\ge 0 $ that is continuous on the interval $[0,1]$ satisfies $\int f_n dm \leq 1$ and strict inequality occur in Fatou's lemma.

I could find one that satisfies the strict inequality but discontinuous. example characteristics function $f_n(x)=\large{X}_{[n,n+1]}(x)$. but I need one that is continuous as well and satisfies all the above. any assisstance is appreciated.

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  • $\begingroup$ Could you take $X_{[n,n+1]}(x)$ and "round out" the edges so that the integral stays less than 1 but it becomes continuous? $\endgroup$ – j4l3kl24jkl2 Nov 11 '16 at 15:28
  • $\begingroup$ By 'round out' the edges you mean make the interval an open one? $(n,n+1)$ . $\endgroup$ – J. Kyei Nov 11 '16 at 15:41
  • $\begingroup$ No I more mean connect the discontinuous points at the edges of the interval in a continuous way such that the integral becomes strictly less than 1. I can post an image if you'd like. $\endgroup$ – j4l3kl24jkl2 Nov 11 '16 at 15:47
  • $\begingroup$ Kindly post the image for me. $\endgroup$ – J. Kyei Nov 11 '16 at 16:00
  • $\begingroup$ Note that those characteristic functions are not defined on $[0,1].$ $\endgroup$ – zhw. Nov 11 '16 at 16:52
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You should be able to make $X_[n,n+1]$ continuous by shrinking the interval where your function is exactly one, and connecting the points in a small epsilon neighborhood of $n$ and $n+1$, like so enter image description here

In this way, the integral is $<1$, but the function becomes continuous.

The "rounding out" that I was referring to was to make the function differentiable, which is not needed in your question. I apologize for misreading, and for my horrible drawing :)

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Try the sequence $f_n(x) = n^2x^n(1-x).$

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  • $\begingroup$ I have 'tested' it against all the conditions and I think it works. $\endgroup$ – J. Kyei Nov 13 '16 at 22:17

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