Prove Riemann Integrability: Let f:[0,1]->R

[mathster]

Let f:[0,1]-> R be bounded on [0,1] and continuous on (0,1). Prove that f is Riemann integrable on [0,1]. Hint: Show that for any epsilon > 0 there exists a partition P of [0,1] such that U(f,P)-L(f,P) < epsilon.

So Let P = {0 = t₀ < t₁ < ... < t_n = 1}
Since its bounded on [0,1] (|f(x)| <= S) and continuous on (0,1) we know its continuous on [t₁,t_n-1]. So Let P1 = { t₁, ..., t_n-1} So its continuous on this interval and bounded we know that it is uniformly continuous and thus integrable on this interval so we know that U(f,P1) - L(f,P1) < epsilon for all epsilon. Now I guess I need to add in the sums of the two intervals on the end to make the partition equal to P. Am I on the right track? SO then I would have U(f,P) - L(f,P) = M(f,[t₀,t₁])*(t₁-t₀) + U(f,P1) + M(f,[t_n-1,t_n])*(t_n-t_n-1) - [ m(f,[t₀,t₁]*(t₁-t₀) + L(f,P1) + m(f,[t_n-1,t_n]*(t_n - t_n-1) ] <= S(t₁-t₀) + U(f,P1) + S(t_n-t_n-1) - [S(t₁-t₀) + L(f,P1) + S(t_n-t_n-1)] = U(f,P1) - L(f,P1) < epsilon.

I think this is true but also feel like I am missing something. Any help is much appreciated.

 

[ThePerfectHacker]

So its continuous on this interval and bounded we know that it is uniformly continuous and thus integrable on this interval so we know that U(f,P1) - L(f,P1) < epsilon for all epsilon.

This is false. Carefully, write the sentence again using the right quantifiers.
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Here is a hint. Suppose $f$ and $g$ are functions defined on $[0,1]$ where $f$ is integrable on $[0,1]$ and $g=f$ for every $x\in [0,1]$ except for finitely many values. Then $g$ is integrable also on this interval.

If $f:[0,1]\to \mathbb{R}$ is bounded consider $\lim_{x\to 0^+}f(x)$. This limit may not necessarily exist, so instead consider $\limsup_{x\to 0^+}f(x)$ which does exist and is equal to some finite number number $a$. In a similar manner let $b = \limsup_{x\to 1^-}f(x)$. Now define function $g = a$ at $x=0$, $g=b$ at $x=1$, and $g=f$ for $0<x<1$. By the above theorem to show that $f$ is integrable on $[0,1]$ it sufficies to show that $g$ is integrable.

Since $\limsup_{x\to 0+} g(x) = g(0) $ for any $\varepsilon > 0$ there is a $0<A < 1$ such that $|g(A) - g(0)| < \varepsilon$. Since $\limsup_{x\to 1^-} g(x) = g(1)$ there is a $A<B<1$ such that $|g(B) - g(1)| < \varepsilon$.

Now on the interval $[A,B]$ the function $g$ is uniformly continuous so we can partition it as $\{ t_1,t_2,...,t_{n-1} \}$ such that $t_1=A,t_{n-1}=B$ and $|g(t_j) - g(t_{j-1})| < \varepsilon$.

Define $P = \{ t_0,t_1,t_2,...,t_{n-1},t_n\}$ such that $t_0=0$ and $t_n = 1$. Then we see that $|g(t_j) - g(t_{j-1}) | < \varepsilon$. From here you should be able to carry out the rest of the argument.

 

[stainburg]

Just take $$P=\{0,1/n,...,i/n,...,1\}$$, $$i=0,2,...,n$$.

Hence, $$f(x)$$ is uniformly continuous on $$[1/n, (n-1)/n]$$. Hence, $$M_i-m_i<\eta$$, for $$1\leq i\leq n-1$$ and large $$n$$.

Let $$M=\sup |f(x)|$$, we get $$M_i-m_i\leq 2M$$.

Therefore

$$U(P,f)-L(P,f)\\=\sum_{1\leq i\leq n-1}(M_i-m_i)/n+(M_0-m_0)/n+(M_n-m_n)/n\leq\eta(n-2)/n+4M/n <\epsilon$$

for large n.