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$$g(x)= \begin{cases} 6 &\text{, if x $\in \mathbb{Q}$}\\ x &\text{, if x $\notin$ $\mathbb{Q}$} \end{cases}$$

If $P$ is a partition: $\{x_0 = 7.1, x_1, x_2, \dots , x_n = 8\}$, find $L(g,P)$ and $U(g,P)$. Simplify sums where possible.

$m_i$ = inf $\{g(x), \text{ all partitions P on [7.1, 8]}\}$

We know that $g$ will take on the value $6$. But we have that $g(x)$ takes on $'x'$ as well, and I'm not sure how to handle this?

1 Answers 1

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Consider a general partition of $[7.1,8]$ that is not necessarily uniform. Since the rationals are dense in any interval, it follows that on each subinterval $[x_{j-1},x_j]$

$$\sup_{x \in [x_{j-1},x_j]}g(x) = \max(x_j,6) = x_j, \\ \inf_{x \in [x_{j-1},x_j]}g(x) = \min(x_{j-1},6) = 6.$$

The upper sum is

$$\begin{align}U(g,P) &= \sum_{j=1}^nx_j(x_j - x_{j-1}) \\ &= \frac{1}{2}\sum_{j=1}^n(x_j+ x_{j-1})(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n(x_j - x_{j-1})(x_j - x_{j-1}) \\ &= \frac{1}{2}\sum_{j=1}^n(x_j^2 - x_{j-1}^2)+ \frac{1}{2}\sum_{j=1}^n(x_j - x_{j-1})^2 \\ &= \frac{x_n^2 - x_0^2}{2 } + \frac{1}{2}\sum_{j=1}^n(x_j - x_{j-1})^2 .\end{align}$$

The second term cannot be further reduced without providing more details on the partition.

Finding the lower sum is straightforward.

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    We have: $\sum_{j=1}^nx_j(x_j - x_{j-1})$, but $\sum_{j=1}^n(x_j - x_{j-1})$ is just sum of all sub-intervals, and thus is basically $8-7.1 = 0.9$. So I think the simplification would be $0.9x_j$? – 2017-01-21
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    @KSplitX: I don't think so. If you are summing over $j$ how can the result stiil depend on $j$, i.e. $0.9 x_j$. I obtained two terms: The first $(x_n^2 - x_0^2)/2$ would be the integral $\int_{x_0}^{x_n} x \, dx $ if the function were Riemann integrable. It is actually the upper integral. The second term that I said cannot be simplified further tends to $0$ as the partition mesh goes to $0$. Do you see what the lower sum is? The supremum of the lower sum (lower integral) will not equal the infimum of the upper sum (upper integral) as this is not Riemann integrable. – 2017-01-21
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    Note that in the problem statement it said "simplify the sums where possible." Hence the second term remains without further simplification. Your logic would be correct for the lower sum $\sum_{j=1}^n6(x_j - x_{j-1})$. – 2017-01-21
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    Because if you assume the interval is $[x_{j-1}, x_j]$, our summation will be as follows: $\sup_{x \in [x_{j-1},x_j]} = (x_{1} - x_{0})+(x_{2}-x_{1})+\dots + (x_{n}+x_{n-1})$ – 2017-01-21
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    Notice how common terms cancel, and we are left with $(x_{n}-x_{0})$ which is $8-7.1 = 0.9$ – 2017-01-21
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    If that does not help, can you explain how your summation suddenly became $\frac{1}{2}$ (second step)? – 2017-01-21
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    It is true that $\sum_{j=1}^n (x_j - x_{j-1}) = x_n -x _0$. But the upper sum is $\sum_{j=1}^n x_j(x_j - x_{j-1})$ which only simplifies as I showed. You are mistaken about how the $\sup$ enters into all of this. It arises in the definition of the upper sum as $U(g,P) = \sum_{j=1}^n \sup_{x \in [x_{j-1},x_j] }g(x)(x_j - x_{j-1})$. The $x_j$ factor in each term is the local supremum of $g$ on the sub-interval. – 2017-01-21
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    My simplification uses $x_j(x_j - x_{j-1}) = (1/2)(x_j + x_{j-1})(x_j - x_{j-1}) + (1/2)(x_j - x_{j-1})(x_j - x_{j-1}) = (1/2)(x_j^2 - x_{j-1}^2) + (1/2)(x_j - x_{j-1})^2$ and then summing. The first sum is telescoping. That is where the $1/2$ factors come from. – 2017-01-21
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    Is this just not becoming clear to you? The complication here is the discontinuity of the function which results in all lower sums taking on the the value $6(x_n - x_0)$ which fails to converge to the same value as the upper integral. Aside from that the treatment of upper sums is identical to the standard evaluation of the integral $\int_a^b x \, dx$ by upper Riemann sums. – 2017-01-22