$f(x)=\cos(x)$ if $x=1/n$ for $n\geqslant1$ and $f(x)=-1$ if not. Show that $f(x)$ is Riemann integrable on$ [0,1]$

analysisintegrationriemann sum

Let $f:[0,1] \rightarrow \mathbb{R}$ such that$f(x)=\cos(x)$ if $x=1/n$ for $n\geqslant1$ and $f(x)=-1$ if not.

So, to show that $f$ is integrable, i need to proof that $\bar{S}_{\sigma}(f) \leqslant \underline{S}_{\sigma}(f) + \epsilon$ $\forall \epsilon > 0$. $\sigma$ denotes a subdivision of $[0,1]$, $\bar{S}$ superior Darboux sum and $\underline{S}$ lower Darboux sum.

I remarked that by density of irrational numbers i can write that $\underline{S}_{\sigma}(f)=-1$ for all subdivision $\sigma$. So, i have to show that

$\bar{S}_{\sigma}(f) +1 \leqslant\epsilon$.

So now i don't really see how to choose correctly the subdivision of $[0,1]$.. If someone could explain it with details i would appreciate it. Thanks in advance.

Best Answer

Fix $\epsilon > 0$.

Let $M=m^2$ where $m > 1$ is an integer such that ${\large{\frac{6}{m}}} < \epsilon$.

Let $\sigma$ be the partition of $[0,1]$ consisting of the $M+1$ intervals $I_1,...,I_{M+1}$ with $I_k=[x_{k-1},x_k]$ where $$ 0=x_0 < x_1 < \cdots < x_{M+1}=1 \;\;\;\;\; $$ and $x_1,...,x_M$ are defined by \begin{align*} x_1&=\frac{1}{m}\\[4pt] x_k&=x_1+(k-1)d,\;\,\text{for}\;2\le k\le M\\[4pt] d&=\frac{1-x_1}{M}=\frac{m-1}{m^3}\\[4pt] \end{align*} If $A,B$ are given by \begin{align*} A&=\left\{{\small{\frac{1}{n}}}{\;{\Large{\mid}}\;}1\le n\le m\right\} \qquad\qquad\;\;\;\, \\[4pt] B&=\left\{{\small{\frac{1}{n}}}{\;{\Large{\mid}}\;}n > m\right\} \\[4pt] \end{align*} then the following claims are immediate:

  • Each element of $B$ lies in the interior of $I_1$.$\\[4pt]$
  • Each element of $A$ lies in at most two of $I_2,...,I_{M+1}$.$\\[4pt]$

Now let $S$ be the Riemann sum given by $$ S=\sum_{k=1}^{M+1}f(x_k^*)\Delta x_k $$ where $x_k^*\in I_k$ and $\Delta x_k=x_k-x_{k-1}$.

Then we get the lower bound \begin{align*} S &= \sum_{k=1}^{M+1}f(x_k^*)\Delta x_k \\[4pt] &\ge \sum_{k=1}^{M+1}(-1)\Delta x_k \\[4pt] &= (-1)\sum_{k=1}^{M+1}\Delta x_k \\[4pt] &= (-1)(1) \\[4pt] &= -1 \end{align*} and since

  • $f(x_k^*)\le 1$ for all $k$.$\\[4pt]$
  • For $2\le k\le M+1$, we have $f(x_k^*)=-1$ with at most $2m$ exceptions.

we get the upper bound \begin{align*} S &= \sum_{k=1}^{M+1}f(x_k^*)\Delta x_k \\[4pt] &= f(x_1^*)\Delta x_1+\sum_{k=2}^{M+1}f(x_k^*)\Delta x_k \\[4pt] &= f(x_1^*)x_1+d\sum_{k=2}^{M+1}f(x_k^*) \\[4pt] &\le (1)(x_1)+d\bigl((1)(2m)+(-1)(M-2m)\bigr) \\[4pt] &= -1+\frac{2(3m-2)}{m^2} \\[4pt] &< -1+\frac{6}{m} \\[4pt] &< -1+\epsilon \\[4pt] \end{align*}

Thus $-1\le S < -1+\epsilon$.

It follows that $f$ is Riemann integrable on $[0,1]$ and ${\displaystyle{\int_0^1 f(x)\,dx = -1}}$.

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