= Solution
{c}
Take $f_n(x)=x^n$ on $[0,1]$. Each <function> is <continuous>, and the <pointwise limit> is $f(x)=0$ for $0\leq x<1$, while $f(1)=1$. Nevertheless
$$
\boxed{\int_0^1 f_n(x)\,dx=\frac1{n+1}\longrightarrow0=\int_0^1 f(x)\,dx.}
$$
The limiting <function> is <Riemann integrable>, since changing a value at one point does not change its <Riemann integral>. There is no <uniform convergence>: for every $n$, $\sup_{[0,1]}|f_n-f|=1$, with the <supremum> approached as $x\uparrow1$ from below. Equivalently, the <uniform limit theorem> would require the limit to be <continuous>.
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