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Background

I am trying to prove the following theorem.

Let $f:[a,b]\to\mathbb{R}$ be a bounded function. If $c\in(a,b)$ then show that $f:[a,b]\to\mathbb{R}$ is Riemann Integrable on $[a,b]$ if $f$ is Riemann Integrable on both $[a,c]$ and $[c,b]$.

Notation

We use the following notations to simplify our discussion.

  • The collection of all partitions on $[a,b]$ for the function $f$ is denoted by $\mathcal{P}(f,[a,b])$.

  • The upper sum for partition $P\in\mathcal{P}(f,[a,b])$ is denoted by $U(f,P,[a,b])$.

  • The lower sum for partition $P\in\mathcal{P}(f,[a,b])$ is denoted by $L(f,P,[a,b])$.

Sketch of the Proof of the Theorem

Observe that, $$(P\in \mathcal{P}(f,[a,c]))\land(Q\in\mathcal{P}(f,[c,b]))\implies P\cup Q\in \mathcal{P}(f,[a,b])\tag{1}$$Furthermore, $$U(f,P\cup Q,[a,b])=U(f,P,[a,c])+U(f,Q,[c,b])\tag{2}$$$$L(f,P\cup Q,[a,b])=L(f,P,[a,c])+L(f,Q,[c,b])\tag{3}$$If we define, $$\mathcal{P}^\ast(f,[a,b]):=\{P\cup Q: (P\in \mathcal{P}(f,[a,c]))\land(Q\in\mathcal{P}(f,[c,b]))\land (P\cap Q=\{c\}) \}$$Then $\mathcal{P}^\ast(f,[a,b])\subseteq \mathcal{P}(f,[a,b])$. Now, observe that, \begin{align}\int_{a}^{\bar{b}}f&=\inf\{U(f,R,[a,b]):R\in\mathcal{P}(f,[a,b])\}\\&\le \inf\{U(f,R,[a,b]):R\in\mathcal{P}^\ast(f,[a,b])\}\\&=\inf\{U(f,P(R),[a,c])+U(f,Q(R),[c,b]):(P(R),Q(R))\in\mathcal{P}(f,[a,c])\times \mathcal{P}(f,[c,b])\}\\&=\inf\{U(f,P,[a,c])+U(f,Q,[c,b]):(P,Q)\in\mathcal{P}(f,[a,c])\times\mathcal{P}(f,[c,b]) \}\\&=\inf\{U(f,P,[a,c]):P\in\mathcal{P}(f,[a,c])\}+\inf\{U(f,Q,[c,b]):Q\in\mathcal{P}(f,[c,b])\}\\&=\int_{a}^c f+\int_{c}^b f\end{align}Similarly, \begin{align}\int_{\bar{a}}^{b}f&=\sup\{L(f,R,[a,b]):R\in\mathcal{P}(f,[a,b])\}\\&\ge \sup\{L(f,R,[a,b]):R\in\mathcal{P}^\ast(f,[a,b])\}\\&=\sup\{L(f,P(R),[a,c])+L(f,Q(R),[c,b]):(P(R),Q(R))\in\mathcal{P}(f,[a,c])\times \mathcal{P}(f,[c,b])\}\\&=\sup\{L(f,P,[a,c])+U(f,Q,[c,b]):(P,Q)\in\mathcal{P}(f,[a,c])\times\mathcal{P}(f,[c,b]) \}\\&=\sup\{L(f,P,[a,c]):P\in\mathcal{P}(f,[a,c])\}+\sup\{L(f,Q,[c,b]):Q\in\mathcal{P}(f,[c,b])\}\\&=\int_{a}^c f+\int_{c}^b f\end{align}Consequently we have, $$\int_{a}^c f+\int_{c}^b f\le \int_{\bar{a}}^{b}f\le \int_{a}^{\bar{b}}f\le \int_{a}^c f+\int_{c}^bf$$which implies, $$\int_{a}^{b}f= \int_{a}^c f+\int_{c}^bf$$

Question

Is my proof correct?

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1 Answer 1

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Your proof is fine, demonstrating both that the integral exists and is given by the sum of integrals over the subintervals. A simpler argument for integrability can be based on the Riemann criterion.

Given any $\epsilon > 0$, there are partitions $Q$ and $Q'$ of $[a,c]$ and $[c,b]$, respectively, such that

$$U(f,Q,[a,c]) - L (f,Q,[a,c]) < \epsilon/2, \\ U(f,Q',[c,b]) - L (f,Q',[c,b]) < \epsilon/2.$$ Let $P = Q \cup Q'$. Using the additivity of Riemann sums, we have for this partition of $[a,b],$

$$U(f,P,[a,b]) - L (f,P,[a,b]) < \epsilon.$$

Therefore, $f$ is integrable on $[a,b]$.

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