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Let $ \ f: [0,1] \to \mathbb{R} \ $ be defined by $$ f(t)=\frac{1}{2^n} , \ \ \ \frac{1}{2^{n+1}}<t<\frac{1}{2^n} \\ = \ 0 \ , \ \ \ \ \ \ \ \ \ \ \ \ \ t=0 $$

Show that $ \ f \in \mathcal{R} [0,1] \ $ (Riemann integrable on $ \ [0,1] )$, by showing that given any $ \ \epsilon >0 \ $ there exists a partition $ \ P \ $ such that $$ U(P,f)-L(P,f) < \epsilon $$

Answer:

Let $ \ \epsilon>0 \ $.

Then since $ \ \frac{1}{2^n} \to 0 \ $ , there exists $ \ N \in \mathbb{N} \ $ such that $ \ \frac{1}{2^n} \in [0, \epsilon] \ \ \forall \ n>N $. Now $ \ [0,1]=[0,\epsilon] \cup [\epsilon,1] \ $ shows that only finite number of $ \ \frac{1}{2^n}' s \ $ lie in the interval $ \ [\epsilon,1] \ $.

So let the parttion be $ \ P=\{0, \epsilon, x_1,x_2,........,x_m \} \ $ , where $ \ x_i \in [\epsilon,1 ] \ $.

This is the partition .

Now my question is how to compute $ \ U(P,f)-L(P,f) \ $ ?

Noa Even
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MAS
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1 Answers1

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You have not define what the $x_1, x_2, \dots ,x_m$ are!

Now for $N \in \mathbb N$, define the partition $$P_N = 0 < \frac{1}{2^N} < \frac{1}{2^{N-1}} < \dots < \frac{1}{2} < 1$$

You have $$U(P_N,f) - L(P_N,f)= \frac{1}{2^{2N}}.$$ So for $\epsilon >0$, the partition $P_N$ will work providing that $\frac{1}{2^{2N}} < \epsilon$.

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