Lebesgue integration

Lebesgue integration builds an integral from measures of measurable sets. On a measure space $(X,\mathcal{A},\mu )$, a nonnegative simple function $s={\sum }_{j=1}^m{a}_j{1}_{E_j}$ on disjoint measurable sets hasHere $1_{E_j}$ is the indicator function of $E_j$, and $0\cdot \infty$ is taken as zero. For any nonnegative measurable function $f$, define its Lebesgue integral by approximation from below:This value may be infinite. For a signed measurable function, write $f=f_{+}-f_{-}$ using its positive part and negative part. A finite Lebesgue integral exists exactly when ${\int }_X|f|d\mu <\infty$; subtract the two finite contributions. One must not subtract two infinite values.^[ref]

For an explicit simple function on $[0,1]$, take $f=2$ on $[0,1/3)$ and $f=5$ on $[1/3,1]$. Its Lebesgue integral is $2(1/3)+5(2/3)=4$: values are weighted by the measures of the sets carrying them. A set of Lebesgue measure zero contributes zero, so changing a function on a null set does not change its Lebesgue integral.

The monotone convergence theorem and dominated convergence theorem make Lebesgue integration useful when passing from approximations to a limit. For the latter, measurable functions $f_n$ converging almost everywhere to $f$ satisfy $\int f_{n}d\mu \to \int fd\mu$ when $|f_n|\le g$ for a fixed integrable function $g$. The common bound prevents a shrinking peak from retaining a nonzero integral while disappearing pointwise.

In the finite, nonzero normalization, let $0<\theta <1$, let $f,g$ be nonnegative integrable functions on ${\mathbb{R}}^n$ with $0<\int f,\int g<\infty$, and let $h$ be a nonnegative measurable function. If they satisfy $h((1-\theta )x+\theta y)\ge f(x{)}^{1-\theta }g(y{)}^{\theta }$ for all $x,y\in {\mathbb{R}}^n$, then their Lebesgue integrals obeyThe quantile derivative identity and monotone substitution inequality prove the one-dimensional case; Tonelli theorem extends it by induction. It implies the Brunn–Minkowski inequality.

For a Lebesgue integrable function $f$ on a measure space $(X,\mathcal{A},\mu )$, small measure forces a small contribution to its Lebesgue integral: for every $\epsilon >0$ there is $\delta >0$ such thatfor every measurable set $E$. The number $\delta$ may depend on $f$ and $\epsilon$, but must work for every such $E$.^[ref]

To prove it, choose $M>0$ with ${\int }_{\{|f|>M\}}|f|d\mu <\epsilon /2$; integrability and the dominated convergence theorem allow this tail to be made arbitrarily small. Split $E$ according to whether $|f|\le M$. Thenwhenever $\mu (E)<\epsilon /(2M)$.

Consequently, if measurable sets $E_n$ satisfy $\mu (E_n)\to 0$, then ${\int }_{E_n}|f|d\mu \to 0$, whether or not the sets are nested. For $f(x)=x^{-1/2}$ on $(0,1)$, for example, the integral on $(0,\delta )$ is $2\sqrt{\delta }$, which still tends to zero despite the unbounded integrand. Merely saying that sets shrink is insufficient: the nested sets $(0,1/2+1/n)\cap (0,1)$ retain a positive limiting measure and a positive integral for $f=1$.