If $f\in L^{+}(E)$, the Lebesgue integral of $f$ is

If $E\subset \mathbb{R}$ is a set with $m(E)=0$, then for all $f\in L^{+}(E)$ we have ${\int }_{E}f=0$

ie it is only interesting to take integrals over functions with positive measure

From the definition, we have $f\in L^{+}(E)$. If $\phi$ is a simple function such that $\phi =\sum _{i=1}^n{a}_i{\chi }_{A_i}$ and $\phi \le f$, then $m(A_i)\le m(A)=0$. So in the sum, all terms must be $0$. Thus we always have ${\int }_E\phi =0$ and the supremum over all such $\phi$ is also $0$.

1. if $\phi \in L^{+}(E)$ is a simple function, then the two definitions of ${\int }_E\phi$ agree
2. If $f,g\in L^{+}(E)$, $c\in [0,\mathrm{\infty }]$, and $f\le g$ on $E$ then

• ${\int }_{E}cf=c{\int }_{E}f$
• ${\int }_{E}f\le {\int }_{E}g$

3. If $f\in L^{+}(E)$ and $F\subset E$ is measurable, then

• ${\int }_{F}f={\int }_{E}f{\chi }_F\le {\int }_{E}f$

If $f,g\in L^{+}(E)$ and $f\le g$ almost everywhere on $E$ then

Let $F=\{x\in E:f(x)\le g(x)\}$.

• This is measurable since $g-f$ is measurable (sums of measurable functions are measurable)
  And $m(F^c)=0$. Then

If $\{f_n{\}}_n$ is a sequence in $L^{+}(E)$ (nonnegative and measurable) such that $f_1\le f_2\le \dots$ and $f_n\to f$ pointwise on $E$, then

Pointwise convergence is much weather than uniform convergence required for Reimann integration

Since $f_1\le f_2\le \dots$, we have that ${\int }_E{f}_1\le {\int }_E{f}_2\le \dots$ (since function relations almost everywhere hold in the integral). Thus ${\left\{{\int }_E{f}_n\right\}}_n$ is an increasing sequence on nonnegative numbers. Thus $\underset{n\to \mathrm{\infty }}{lim}{\int }_E{f}_n\in [0,\mathrm{\infty }]$.

And since $\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f(x)$ for all $x\in E$, we know $f_1\le f_2\le \cdots \le f$

Thus for all $n$, we have

Thus to show equality, we have need to show ${\int }_{E}f\le \underset{n\to \mathrm{\infty }}{lim}{\int }_E{f}_n$.

Now, let $\phi \in L^{+}(E)$ be simple ie $\phi =\sum _{i=1}^m{a}_i{\chi }_{A_i}$ with $\phi \le f$. Let $ϵ\in (0,1)$. and define

Note that for all $x$, we then have $(1-ϵ)\phi (x)<f(x)$. And since $\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f$, every $x$ must lie in some $E_n$. Thus $\bigcup _n{E}_n=E$

Then we have

And since $E_1\cap A_i\subset E_2\cap A_i\subset \dots$ and $\bigcup _n(E_n\cap A_i)=A_i$, this implies that

Thus

Thus for all $ϵ\in (0,1)$ we have

Let $f\in L^{+}(E)$ and let $\{{\phi }_n{\}}_n$ be a sequence of simple functions such that

with ${\phi }_n\to f$ pointwise. Then

ie, we can take any pointwise increasing sequence of simple functions and compute the limit (instead of needed to compute the supremum)

We can just plug the ${\phi }_n$ in to the Monotone Convergence Theorem

If $f,g\in L^{+}(E)$ then

Let $\{{\phi }_n{\}}_n$ and $\{{\psi }_n{\}}_n$ be two sequences of simple functions such that
$0\le {\phi }_1\le {\phi }_2\le \cdots \le f$ with ${\phi }_n\to f$ and
$o\le {\psi }_1\le {\psi }_2\le \cdots \le g$ with ${\psi }_n\to g$ pointwise then

Where ${\phi }_n+{\psi }_n\to f+g$ pointwise and each ${\phi }_i+{\psi }_i$ are simple (since they are the sum of simple functions). So by the Monotone Convergence Theorem and linearity for simple functions we have

Let $\{f_n{\}}_n$ be a sequence in $L^{+}(E)$. Then

via induction. Since the lebesgue integral of sum is sum of the integral, we know for each $N\in \mathbb{N}$ that

And since

and $\sum _{n=1}^N{f}_n\to \sum _{n=1}^{\mathrm{\infty }}{f}_n$ as $N\to \mathrm{\infty }$, by the Monotone Convergence Theorem we have

this does not hold for Riemann integration. Enumerate the rationals and let $f_n$ be the function that is 1 for the first $n$ rational numbers and 0 everywhere else

Let $f\in L^{+}(E)$. Then ${\int }_{E}f=0⟺f=0$ almost everywhere on $E$

If $\{f_n{\}}_n$ is a sequence in $L^{+}(E)$ such that $f_1(x)\le f_2(x)\le \dots$ for almost all $x\in E$ and

Then

Let

Then $m(E\setminus F)=0$ by assumption. Thus $f-{\chi }_{F}f=0$ a.e. and $f_n-{\chi }_F{f}_n=0$ a.e. also. Then the Monotone Convergence Theorem says

Where the first equality holds since function relations almost everywhere hold in the integral and the last holds by the Monotone Convergence Theorem. This then becomes

Because $m(E\setminus F)=0$ and so any integral over the region is 0.

ie, sets of measure zero do not affect the Lebesgue Integral.

Let $\{f_n{\}}_n$ be a sequence in $L^{+}(E)$. Then

We have

And since $\underset{k\ge 1}{inf}{f}_k(x)\le \underset{k\ge 2}{inf}{f}_k\le \dots$ by the MCT we have

For all $j\ge n$ and all $x\in E$, we have

So we have "swapped the integral and inf" and we can plug this into the MCT to get the desired result.

Let $f\in L^{+}(E)$ and let ${\int }_{E}f<\mathrm{\infty }$. Then the set

must have measure 0.

We know for all $n\in \mathbb{N}$ we have

So integrating both sides gives us

Thus for all $n$, we have $m(F)\le \frac{1}{n}{\int }_{E}f$. Taking $n\to \mathrm{\infty }$, we have $m(F)=0$.