Verify the statements for Riemann-integrable function $f_n(x)$. For each $n = 1, 2, \cdots$ a function $f_n(x)$ is defined so that it is Riemann-integrable on $[a, b]$ and the series $\sum_{n=1}^{\infty}f_n(x)$ converges $\forall \space x \in [a,b]$. 
Which of the following statements are true?
$$\lim_{n\rightarrow\infty}\sup_{x\in[a,b]}|f_n(x)| = 0$$
$$\int_{a}^{b}\bigg(\lim_{n\rightarrow \infty} |f_n(x)|\bigg) dx = 0$$
$$\lim_{n\rightarrow\infty}\int_{a}^{b}|f_n(x)|dx = 0$$
ATTEMPT
I. $f_n$ is integrable, so it is bounded. Therefore, $\sup |f_n(x)|$ exists and equals to $|f_n(c)|, c\in [a, b]$. Since the series converges, $\lim f_n(x) = 0 \space \forall x \in [a, b]$. Hence $\lim \sup _{x \in [a, b]}|f_n(x)| =\lim|f_n(c)| = |0| = 0.$ Answers indicate that this is false, but I don't see where I am mistaken.
II. Recall that $\lim f_n(x) = 0 \space \forall x \in [a, b]$. Then $\lim |f_n(x)| = |0| = 0 \space \forall x \in [a, b]$. Therefore, the integrand is 0 on $[a, b]$, and so is the integral.
III. No idea. 
It would be great if there was a counterexample.

Let $ f_n$ defined at $ [0,1]$ by
$$f_n(x)=x^n \; if \; x\ne 1 \; and \; f_n(1)=0$$
$$f_n \;  is \; Riemann\;  integrable\; at \; [0,1],$$
$$\sum f_n(x) \; converges \; \forall x\in[0,1]$$
but
$$\sup_{x\in [0,1]}|f_n(x)|=1$$
the first statement is false.
