# proof that a function is integrable on a interval $[a,b]$

a) Divide a interval $[a,b]$ into $n$ equal subintervals.

• here I'm thinking $P_{n} =(x_0,x_1,x_2,x_3,x_{n-1}, x_n)$ where $a = x_0 < x_1 < x_2 < x_3 <\dots< x_{n-1} < x_n = b$

b) make an expression for the lower Riemann sum $L(f,P_{n})$ and the upper $U(f,P_{n})$.

• here I also got an idea, what to do. The hardest part is the last 4 assignments:

c) show that $U(f,P_n) - L(f,P_n) = \frac{(b-a)(f(b)-f(a))}{n}$

d) show that for a given $\epsilon>0$ calculate a $n'$ so that $U(f,P_{n'}) - L(f,P_{n'}) < \epsilon$

e) use the result from d) to show that there is only one number $I$ which obeys $U(f,P_n) ≥ I ≥ L(f,P_n) \quad\forall n \in \mathbb{N}$

f) Show that $f(x)$ is integrable on $[a,b]$
Where do you use that $f$ is continuous (if you demand this at all)?

I don't know what to do what the last 4 assigments.

Kind regards Jones

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c is false. For example, take $a=0$, $b=1$, $f(x)=x-x^2$. You have probably left out a very important assumption: That $f$ is non-decreasing. –  Harald Hanche-Olsen Jan 26 '13 at 12:02
For a you are asked to make equal subintervals. That specifies the $x_i$ completely. For b you (of course) need knowledge about $f$ as it is not possible for every function $f$ to prove that it is integrable. –  Hagen von Eitzen Jan 26 '13 at 12:04

As you are asked to specifically use equal subintervals, you can write explicitly that $x_i=a+i\frac{b-a}n$, $0\le i\le n$.
By definition, $$L(f,P_n)=\sum_{i=1}^n(x_i-x_{i-1})\inf\{f(x)\mid x_{i-1}\le x\le x_i\}$$ and $$U(f,P_n)=\sum_{i=1}^n(x_i-x_{i-1})\sup\{f(x)\mid x_{i-1}\le x\le x_i\}.$$ Assuming that the function $f$ is non-decreasing (a property of $f$ that is not mentioned in your problem statement, but matches well the expected results), we know that $\inf\{f(x)\mid x_{i-1}\le x\le x_i\}=f(x_{i-1})$ and $\sup\{f(x)\mid x_{i-1}\le x\le x_i\}=f(x_i)$. Additionally using $x_i-x_{i-1}=\frac{b-a}n$, we obtain \begin{align}U(f,P_n)-L(f,P_n)&=\frac{b-a}n\sum_{i=1}^nf(x_i)-\frac{b-a}n\sum_{i=1}^nf(x_{i-1})\\&=\frac{b-a}n\sum_{i=1}^nf(x_i)-\frac{b-a}n\sum_{i=0}^{n-1}f(x_{i})\\&=\frac{b-a}n(f(x_n)-f(x_0))=\frac{(b-a)(f(b)-f(a))}{n}\end{align}.
In the light of c it suffices to choose $n'>\frac{(b-a)(f(b)-f(a))}{\epsilon}$.
As the existence of at least one such $I$ is not asked for, we need only assume that there exist two such numbers $I_1< I_2$ and find a contradiction by letting $\epsilon=I_2-I_1$ in d.
Your definitin of integrability should be something like a restatement of e (however, does it really make use only of equidistant partitions?). We did not make use of continuity, but only that $f$ is non-decreasing. (To repeat: I assume that the fullproblem statement includes this property of $f$)
We might add: A picture makes the equality in c immediately obvious: Note that $U(f,P_n)-L(f,P_n)=\sum_{i=1}^n\bigl(f(x_i)-f(x_{i-1})\bigr)\cdot(b-a)/n$. Think of the $i$th term of the sum as the area of a small rectangle with opposite corners at $\bigl(x_{i-1},f(x_{i-1})\bigr)$ and $\bigl(x_i,f(x_i)\bigr)$. Draw a picture of this! Then note that you can slide these small rectangles horizontally, stacking them on top of each other to form a rectangle $f(b)-f(a)$ tall and $(b-a)/n$ wide. –  Harald Hanche-Olsen Jan 26 '13 at 13:21