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Let $\{f_n\}$ be a sequence of measurable $\&$ continuous functions from $[0,1]$ to $[0,1]$. Assume $f_n \rightarrow f$ pointwise. Is it true/false that,

  1. $f$ is Riemann integrable $\& \int _{[0,1]}f_n \rightarrow \int_{[0,1]}f$?

  2. $f$ is Lebesgue integrable $\& \int _{[0,1]}f_n \rightarrow \int_{[0,1]}f$?

My work:

For $(1),$ I came up with a counter-example $f_n(x)=nx(1-x^2)^n$ on $[0,1]$ as $f_n \rightarrow 0$ but $\int _{[0,1]}f_n=\frac12$ (but is $f_n(x) \in [0,1]$ for all $x$??)

For $(2),$ I think this holds since,

1.The continuous functions over a closed bounded interval is R.I & hence L.I

2.measure space is finite.

  1. $\{f_n\}$ is uniformly bounded and it converges to $f$ pointwise.

So,by DCT, this holds true.

Am I correct? Also what about my choice of function for case (1)?

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For 2) your answer is correct. But for 1) the question is really whether $f$ has to be RI. Once it is RI then the convergence of the integrals will follow from 2) becasue Riemann integral coincides with Lebesgue integral.

To construct a counter-example for 1) arrange the rational numbers in $[0,1]$ in a sequnce $(r_n)$. Let $f_n$ have the value $1$ at $r_1,r_2,...r_n$ and $0$ outside that intervals $(r_i-\frac 1n, r_i+\frac 1 n), 1\leq i \leq n$. You can construct a piece-wise linear function with these properties such that $0 \leq f_n \leq 1$ . Then $f(x)=\lim f_n(x)=1$ for $x$ rational and $f(x)=0$ for $x$ irrational. This $f$ is not be RI.

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  • $\begingroup$ $f$ is not R.I because of the Lebesgue criterion for the Riemann integrability? $\endgroup$ – SL_MathGuy Jan 2 at 2:07
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    $\begingroup$ @SL_MathGuy Yes,but you can also write down the upper and lower Riemann sums and see that $f$ is not RI directly from the definition. $\endgroup$ – Kavi Rama Murthy Jan 2 at 5:22
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$$ \text{Let } f_n(x) = \begin{cases} 2n\left(1-|2nx-1| \right) & \text{if } 0\le x\le1/n, \\[6pt] {} 0 & \text{otherwise.} \end{cases} $$ Then $$ \lim_{n\to\infty} \int\limits_{[0,1]} f_n(x)\,dx = 1 \ne 0 = \int\limits_{[0,1]} \lim_{n\to\infty} f_n(x)\,dx. $$

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  • $\begingroup$ Your sequence of functions aren't continuous $\endgroup$ – K.Power Jan 2 at 1:07
  • $\begingroup$ @K.Power : ok, I've modified them somewhat. $\endgroup$ – Michael Hardy Jan 2 at 1:28

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