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Following is theorem 6.15 of baby Rudin:

If $a<s<b$, $f$ is bounded on $[a,b]$. $f$ is continuous at $s$, then $\alpha(x) = I(x-s)$, then $\int_a^b f d \alpha = f(s)$. $\alpha(x)= I(x-s)$ is the unit step function, $\alpha= 0$ if $x \le s ,\alpha= 1$ if $x >s.$

Proof: Consider partitions $P = \{x_0,x_1,x_2,x_3 \}$, where $x_0 = a, x_1=s, x_2< x_3=b$. Then $U(P,f, \alpha) = M_2, L(P,f, \alpha)=m_2$. Since $f$ is continuous at $s$, we see that $M_2$ and $m_2$ converge to $f(s)$ as $x_2 \to s$. $\square$

Need some help on shedding some light on the bold sentence, especially on how it is used to show that $\int_a^b f d \alpha = f(s)$. I calculated that the difference between upper sum and lower sum is $M_2 - m_2$, but then I am stuck and I do not get the last sentence of the proof.

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Just for completeness, you should define $I$. Presumably, it is the step function? –  Thomas Andrews Jan 14 '13 at 21:29
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Given $ \epsilon > 0$ there exists $\delta > 0$, such that $|x - x_2| < \delta $ implies $f(x_2) - c < f(x) < f(x_2) + \epsilon$ hence $\sup_{ x \in (x_2 - \delta, x_2 + \delta) } f(x)$ and $\inf_{ x \in (x_2 - \delta, x_2 + \delta )} f(x) $ both $\in [ f(x_2) - \epsilon , f(x_2) + \epsilon ]$, now consider what happens when $x_3 - x_2 < \delta.$ –  ACARCHAU Jan 14 '13 at 21:31

2 Answers 2

up vote 1 down vote accepted

Because $f$ is continuous at $s$, given $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|s - t| < \delta$ then $|f(s) - f(t)| < \varepsilon/2$. When Rudin says "as $x_2 \rightarrow s$," he means "chosing a partition $P$ such that $x_2 - x_1 < \delta$," because then for all $t \in [x_1, x_2]$, $|f(s) - f(t)| < \varepsilon/2$, so certainly $M_2 - f(s) < \varepsilon$ and $f(s) - m_2 < \varepsilon$. Then $M_2 - m_2 < 2\varepsilon$, so $f$ is integrable.

Furthermore, $m_2 = L(P, f, \alpha) \le \int_a^b f \hspace{0.03 in} d \alpha \le U(P,f, \alpha) = M_2$, so the above implies that $$f(s) - \varepsilon < m_2 \le \int_a^b f \hspace{0.03 in} d \alpha \le M_2 < f(s) + \varepsilon$$ $$|f(s) -\int_a^b f \hspace{0.03 in} d \alpha| < \varepsilon,$$

Which he have shown for arbitrary $\varepsilon$, so $\int_a^b f \hspace{0.03 in} d \alpha = f(s)$.

Edit: now require $|f(s) - f(t)| < \varepsilon/2$ because $f$ does not necessarily take on its supremum or infimum on $[x_1, x_2]$, as proximal noted in his answer.

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Let's expand Rudin's (admittedly sparse) notation.

$$U(P,f,\alpha) = \sum_{i=1}^3 \sup_{[x_{i-1},x_i]} f(x) \Delta\alpha_i = \sup_{[x_1,x_2]}f(x) = M_2,$$

$$U(P,f,\alpha) = \sum_{i=1}^3 \inf_{[x_{i-1},x_i]} f(x) \Delta\alpha_i = \inf_{[x_1,x_2]}f(x) = m_2.$$

Since $x_1=s$, we have

$$M_2 = \sup_{[s,x_2]}f(x),$$ $$m_2 = \inf_{[s,x_2]}f(x).$$

By continuity at $s$, given $\varepsilon > 0$ there exists $\delta > 0$ such that $|x_2-s|<\delta$ implies $|f(x_2)-f(s)|<\varepsilon/2$. Taking $y \in [s,x_2]$ such that $$\left|\sup_{[s,x_2]}f(x) - f(y)\right| < \varepsilon/2,$$ we get by the triangle inequality

$$\left|\sup_{[s,x_2]}f(x) - f(s)\right| \leq \left|\sup_{[s,x_2]}f(x) - f(y)\right|+\left|f(y) - f(s)\right|< \varepsilon/2+\varepsilon/2 = \varepsilon.$$

This shows that $M_2 \to f(s)$ as $x_2 \to s$. The same argument for $\inf$ shows that $m_2\to f(s)$ as well. Since $M_2$ and $m_2$ tend to the same limit, their difference tends to $0$.

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Yet another example of how if anyone other than Walter Rudin wrote Principles it would have ended up three times as long. –  kigen Jan 14 '13 at 22:11

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