\b[No.] Take $f_n=\mathbf1_{[n,n+1]}$ on the <real line>. At every fixed point it eventually vanishes, so $f=0$, while $\int f_n=1$ for every $n$. Thus the limit of the integrals is one and the integral of the pointwise limit is zero. Nonnegativity alone cannot replace monotonicity or an <integrable> domination hypothesis.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D / 3 / a / ii / Solution

 Solution (source code)