# The Unapologetic Mathematician

## More Properties of Integrals

Today we will show more properties of integrals 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 to this post once we establish the same basic linearity and order properties for the integrals of wider and wider classes of functions.

First up: if $f$ and $g$ are integrable simple functions with $f\geq g$ a.e. then $\displaystyle\int f\,d\mu\geq\int g\,d\mu$

Indeed, the function $f-g$ is nonnegative a.e., and so we conclude that $\displaystyle0\leq\int f-g\,d\mu=\int f\,d\mu-\int g\,d\mu$

Next, if $f$ and $g$ are integrable simple functions then $\displaystyle\int\lvert f+g\rvert\,d\mu\leq\int\lvert f\rvert\,d\mu+\int\lvert g\rvert\,d\mu$

Here we use the triangle inequality $\lvert f+g\rvert\leq\lvert f\rvert+\lvert g\rvert$ and invoke the previous result.

Now, if $f$ is an integrable simple function then $\displaystyle\left\lvert\int f\,d\mu\right\rvert\leq\int\lvert f\rvert\,d\mu$

The absolute value $\lvert f\rvert$ is greater than both $f$ and $-f$, and so we find \displaystyle\begin{aligned}\int f\,d\mu\leq&\int\lvert f\rvert\,d\mu\\-\int f\,d\mu\leq&\int\lvert f\rvert\,d\mu\end{aligned}

which implies the inequality we asserted.

As a heuristic, this last result is sort of like the triangle inequality to the extent that the integral is like a sum; adding inside the absolute value gives a smaller result than adding outside the absolute value. However, we have to be careful here; the integral we’re working with is not the limit of a sum like the Riemann integral was. In fact, we have no reason yet to believe that this integral and the Riemann integral have all that much to do with each other. But that shouldn’t stop us from using this analogy to remember the result.

Finally, if $f$ is an integrable simple function, $E$ is a measurable set, and $\alpha$ and $\beta$ are real numbers so that $\alpha\leq f(x)\leq\beta$ for almost all $x\in E$, then $\displaystyle\alpha\mu(E)\leq\int\limits_Ef\,d\mu\leq\beta\mu(E)$

Indeed, the assumed inequality is equivalent to the assertion that $\alpha\chi_E\leq f\chi_E\leq\beta\chi_E$ a.e., and so — as long as $\mu(E)<\infty$ — we conclude that $\displaystyle\int\alpha\chi_E\,d\mu\leq\int f\chi_E\,d\mu\leq\int\beta\chi_E\,d\mu$

which is equivalent to the above. On the other hand, if $\mu(E)=\infty$, then $f$ must be zero on all but a portion of $E$ of finite measure or else it wouldn’t be integrable. Thus, in order for the assumed inequalities to hold, we must have $\alpha\leq0$ and $\beta\geq0$. The asserted inequalities are then all but tautological.

May 26, 2010 Posted by | Analysis, Measure Theory | 5 Comments