# Sequence of measurable $\&$ continuous functions defined on $[0,1]$

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)?

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.
• $f$ is not R.I because of the Lebesgue criterion for the Riemann integrability? – SL_MathGuy Jan 2 at 2:07
• @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. – Kavi Rama Murthy Jan 2 at 5:22
$$\text{Let } f_n(x) = \begin{cases} 2n\left(1-|2nx-1| \right) & \text{if } 0\le x\le1/n, \\[6pt] {} 0 & \text{otherwise.} \end{cases}$$ Then $$\lim_{n\to\infty} \int\limits_{[0,1]} f_n(x)\,dx = 1 \ne 0 = \int\limits_{[0,1]} \lim_{n\to\infty} f_n(x)\,dx.$$