Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 25J (Probability and Measure)

An $\mathcal{A}$-simple function is a measurable function taking finitely many values, written $s={\sum }_{j=1}^m{a}_j{1}_{A_j}$ with disjoint measurable $A_j$. For nonnegative simple functions define $\int sd\mu ={\sum }_j{a}_j\mu (A_j)$, with $0\cdot \infty =0$. Refining two disjoint descriptions shows this value is independent of the description. For nonnegative measurable $h$, defineFor a real measurable $f$, put $f^{+}=\max (f,0)$ and $f^{-}=\max (-f,0)$. It is Lebesgue integrable when $\int |f|d\mu <\infty$, equivalently both positive and negative parts have finite integrals. ThenThis includes Lebesgue integrable signed simple functions, whose integral is the same finite weighted sum.

Take the truncated dyadic approximationIt has finitely many values and measurable level sets, so is simple. Downward rounding and truncation ensure $0\le g_n\le f$. For a fixed finite $f(x)$, eventually $n>f(x)$, and then $0\le f(x)-g_n(x)<2^{-n}$. Thus $g_n(x)\to f(x)$ everywhere. If extended value infinity is permitted, define $g_n=n$ there, also giving the required limit.

For a fixed $c>0$, the functions $|f_n|1_{\{|f_n|\le c\}}$ tend pointwise to zero and are dominated by the Lebesgue integrable constant $c$, since $\mu (\Omega )<\infty$. The dominated convergence theorem therefore gives convergence of their integrals to zero. SplitTake the upper limit in $n$ and then send $c\to \infty$. The specified uniform integrability condition makes the right side zero. It also ensures integrability of each $f_n$, by the same split for a sufficiently large $c$. Hence $\int |f_n|d\mu \to 0$, and in particular $\int f_{n}d\mu \to 0$.

For a counterexample without the tail condition, use Lebesgue measure on $(0,1)$ and $f_n(x)=n{1}_{(0,1/n)}(x)$. Every fixed positive $x$ eventually lies outside the shrinking interval, so $f_n(x)\to 0$, but ${\int }_0^1{f}_{n}dx=1$ for every $n$.

Fatou's lemma states that nonnegative measurable functions satisfyTo prove it, put $h_n={\inf }_{k\ge n}{f}_k$. These measurable nonnegative functions increase to $\mathrm{liminf}{f}_n$. By the monotone convergence theorem, $\int \mathrm{liminf}{f}_n={\lim }_n\int h_n$. Since $h_n\le f_k$ for every $k\ge n$, one has $\int h_n\le {\inf }_{k\ge n}\int f_k$. Taking the limit proves the inequality, including extended infinite integrals.