Prove that continuous functions are integrable.

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Problem: 

Prove that continuous functions are integrable. Prove that if $f(x)$ is continuous on the closed interval $[a,b]$, then $f(x)$ is integrable there.

Answer: 

It is true that continuous functions are integrable.

SAMPLE SOLUTION

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Solution: 

Recall that if $f(x)$ is continuous on the closed interval $[a,b]$, then $f(x)$ is uniformly continuous on $[a,b]$. So for this problem we can say that for every <Sign in to see all the formulas> there is a <Sign in to see all the formulas> such that for any $z$ and $z'$ in $[a,b]$ if <Sign in to see all the formulas>, then <Sign in to see all the formulas>.

Now let's assume that we have a partition $P$ such that <Sign in to see all the formulas> (remember that $||P||$ is the partition's norm). So let's focus our attention on an arbitrary interval of the partition and denote it by <Sign in to see all the formulas>. Now because $f(x)$ is continuous on the interval <Sign in to see all the formulas> we know from the Extreme-Value Theorem that there are two points $e_i$ and $E_i$ in <Sign in to see all the formulas> where $f(x)$ will take in its minimum and maximum values which we denote by $m_i$ and $M_i$, respectively. In other words, <Sign in to see all the formulas> and <Sign in to see all the formulas>.

Note that <Sign in to see all the formulas>. And because $f(x)$ is uniformly continuous as discussed above, we know that <Sign in to see all the formulas>. Therefore, <Sign in to see all the formulas>.

So now let's form our upper and lower sums and subtract to get

<Sign in to see all the formulas>


But because we know that <Sign in to see all the formulas> we have

<Sign in to see all the formulas>

Recall from this lesson that one of our three equivalent definitions of the definite integral is the following: for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that for every partition $P$, if <Sign in to see all the formulas>, then <Sign in to see all the formulas>. But we have just proven this! Therefore, $f(x)$ is integrable. The proof is now complete.