5.2 Integrable functions  (Page 3/4)

On the other hand, all the definitions of integrability on $[a,b]$ include among the integrable functions the continuous ones. And, all the different definitions of integral give the same value to a continuous function.The differences then in these definitions shows up at the point of sayingexactly which functions are integrable. Perhaps the most enlightening thing to say in this connection is thatit is impossible to make a “good” definition of integrability in such a way that every function is integrable.Subtle points in set theory arise in such attempts, and many fascinating and deep mathematical ideas have come from them.However, we will stick with our definition, since it is simpler than Riemann's and is completely sufficient for our purposes.

Let $[a,b]$ be a fixed closed and bounded interval, and let $I\left(\right[a,b\left]\right)$ denote the set of integrable functions on $[a,b].$ Then:

1. Every element of $I\left(\right[a,b\left]\right)$ is a bounded function. That is, integrable functions are necessarily bounded functions.
2.   $I\left(\right[a,b\left]\right)$ is a vector space of functions.
3.   $I\left(\right[a,b\left]\right)$ is closed under multiplication; i.e., if $f$ and $g\in I\left(\right[a,b\left]\right),$ then $fg\in I\left(\right[a,b\left]\right).$
4. Every step function is in $I\left(\right[a,b\left]\right).$
5. If $f$ is a continuous real-valued function on $[a,b],$ then $f$ is in $I\left(\right[a,b\left]\right).$ That is, every continuous real-valued function on $[a,b]$ is integrable on $[a,b].$

Let $f\in I\left(\right[a,b\left]\right),$ and write $f=lim{h}_n,$ where $\left\{h_n\right\}$ is a sequence of step functions that converges uniformly to $f.$ Given the positive number $ϵ=1,$ choose $N$ so that $|f\left(x\right)-h_N\left(x\right)|<1$ for all $x\in [a,b].$ Then $\left|f\left(x\right)\right|\le |h_N\left(x\right)|+1$ for all $x\in [a,b].$ Because $h_N$ is a step function, its range is a finite set, so that there exists a number $M$ for which $|h_N\left(x\right)|\le M$ for all $x\in [a,b].$ Hence, $\left|f\right(x\left)\right|\le M+1$ for all $x\in [a,b],$ and this proves part (1).

Next, let $f$ and $g$ be integrable, and write $f=lim{h}_n$ and $g=lim{k}_n,$ where $\left\{h_n\right\}$ and $\left\{k_n\right\}$ are sequences of step functions that converge uniformly to $f$ and $g$ respectively. If $s$ and $t$ are real numbers, then the sequence $\{s{h}_n+t{k}_n\}$ converges uniformly to the function $sf+tg.$ See parts (c) and (d) of [link] . Therefore, $sf+tg\in I\left(\right[a,b\left]\right),$ and $I\left(\right[a,b\left]\right)$ is a vector space, proving part (2).

Note that part (3) does not follow immediately from [link] ; the product of uniformly convergent sequences may not be uniformly convergent.To see it for this case, let $f=lim{h}_n$ and $g=lim{k}_n$ be elements of $I\left(\right[a,b\left]\right).$ By part (1), both $f$ and $g$ are bounded, and we write $M_f$ and $M_g$ for numbers that satisfy $\left|f\left(x\right)\right|\le M_f$ and $\left|g\left(x\right)\right|\le M_g$ for all $x\in [a,b].$ Because the sequence $\left\{k_n\right\}$ converges uniformly to $g,$ there exists an $N$ such that if $n\ge N$ we have $|g\left(x\right)-k_n\left(x\right)|<1$ for all $x\in [a,b].$ This implies that, if $n\ge N,$ then $|k_n\left(x\right)|\le M_g+1$ for all $x\in [a,b].$

Now we show that $fg$ is the uniform limit of the sequence $h_n{k}_n.$ For, if $n\ge N,$ then

which implies that $fg=lim\left(h_n{k}_n\right).$

If $h$ is itself a step function, then it is obviously the uniform limit of the constant sequence $\left\{h\right\},$ which implies that $h$ is integrable.

Finally, if $f$ is continuous on $[a,b],$ it follows from [link] that $f$ is the uniform limit of a sequence of step functions, whence $f\in I\left(\right[a,b\left]\right).$