# The Unapologetic Mathematician

## Basic Properties of Integrable Simple Functions

We want to nail down a few basic properties of integrable simple functions. We define two simple functions $f\sum\alpha_i\chi_{E_i}$ and $g\sum\beta_j\chi_{F_j}$ to work with.

First of all, if $f$ is simple then the scalar multiple $\alpha f$ is simple for any real number $\alpha$. Indeed, $\alpha f=\sum\alpha\alpha_i\chi_{E_i}$, and so the exact same sets $E_i$ must have finite measure for both $f$ and $\alpha f$ to be integrable. It’s similarly easy to see that if $f$ and $g$ are both integrable, then $f+g$ is integrable. Thus the integrable simple functions form a linear subspace of the vector space of all simple functions.

Now if $f$ is integrable then the product $fg$ is integrable whether or not $g$ is. We can write

$\displaystyle fg=\sum\limits_{i,j}\alpha_i\beta_j\chi_{E_i}\chi_{F_j}=\sum\limits_{i,j}\alpha_i\beta_j\chi_{E_i\cap F_j}$

If each $E_i$ has finite measure, then so does each $E_i\cap F_j$, whether each $F_j$ does or not. Thus we see that the integrable simple functions form an ideal of the algebra of all simple functions.

We can use this to define the integral of a function over some range other than all of $X$. If $f$ is an integrable simple function and $E$ is a measurable set, then $f\chi_E$ is again an integrable simple function. We define the integral of $f$ over $E$ as

$\displaystyle\int\limits_Ef\,d\mu=\int f\chi_E\,d\mu$

This has the effect of leaving $f$ the same on $E$ and zeroing it out away from $E$. Thus the integral over the rest of the space $E^c$ contributes nothing.

The next two properties are easy to prove for integrable simple functions, but they’re powerful. Other properties of integration will be proven in terms of these properties, and so when we widen the class of functions under consideration we’ll just have to reprove these two. The ones we will soon consider will immediately have proofs parallel to those for simple functions.

Not only is the function $\alpha f+\beta g$ integrable, but we know its integral:

$\displaystyle\int\alpha f+\beta g\,d\mu=\alpha\int f\,d\mu+\beta\int g\,d\mu$

Indeed, if you were paying attention yesterday you’d have noticed that we said we wanted integration to be linear, but we never really showed that it was. But it’s not really complicated: the expression $\sum(\alpha\alpha_i)\chi_{E_i}+\sum(\beta\beta_j)\chi_{F_j}$ represents $f+g$ as a simple function, and it’s clear that the formula holds as stated.

Almost as obvious is the fact that if $f$ is nonnegative a.e., then $\int f\,d\mu\geq0$. Indeed in any representation of $f$ as a simple function, any term $E_i$ corresponding to a negative $\alpha_i$ must have $\mu(E_i)=0$ or else $f$ wouldn’t be nonnegative almost everywhere. But then the term $\alpha_i\mu(E_i)$ contributes nothing to the integral! Every other term has a nonnegative $\alpha_i$ and a nonnegative measure $\mu(E_i)$, and thus every term in the integral is nonnegative. This is the basis for all the nice order properties we will find for the integral.

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Posted by John Armstrong | Analysis, Measure Theory

## 4 Comments	»

1. [...] of simple functions. But the neat thing is that they will follow from the last two properties we showed yesterday. And so their proofs really have nothing to do with simple functions. We will be able to point back [...]

Pingback by | May 26, 2010 | Reply

2. A ‘g’ in the final integral?

Comment by Hunt | May 30, 2010 | Reply

3. Thanks, copy/paste error.

Comment by | May 30, 2010 | Reply

4. [...] of all, from what we know about convergence in measure and algebraic and order properties of integrals of simple functions, we can see that if and are integrable [...]

Pingback by | June 3, 2010 | Reply

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