Distortion of Riemann Integrable function at finite number of points makes it Riemann Integrable again

real-analysisriemann-integrationsolution-verification

I am trying to prove following theorem.Is there any flaw in following proof

consider $f:[a,b] \rightarrow \mathbb{R}$ and $f$ is Riemann
Integrable.Now consider $\bar{f} : [a,b] \rightarrow \mathbb{R}$
differing only at finite number of points from $f$

i)Prove that $\bar{f}$ is Riemann Integrable

ii)$\int_{a}^{b} f = \int_{a}^{b} \bar{f}$

outline of proof is as follows :

  • prove if functions differ at one point and conclude by induction

  • show that $U(f)$ (upper integral of original function) is infimum of the set $\{U(\bar{f},P) : P \in \mathcal{P}\}$ and since infimum is unique, $U(f) = U(\bar{f})$

  • show that $L(f)$ (lower integral of original function) is supremum of the set $\{L(\bar{f},P): P \in \mathcal{P}\}$ and since supremum is unique $L(f) = L(\bar{f})$

  • since $f$ is integrable we will have that $L(\bar{f}) = U(\bar{f})$ proving the both results.

let $x_0$ be the point at which they differ and $D = |\bar{f}(x_0) – f(x_0)|$

first note that since $U(f) = \inf \{ U(f,P) : P \in \mathcal{P}\}$ there exists $P_{0}$ s.t $U(f) > U(f,P_{0}) – \epsilon/2$ now consider refinement of parition $P_{0}$ as $P' = P_{0} \,\cup \{x- \frac{\epsilon}{4D} , x + \frac{\epsilon}{4D} \}$.but we know that $U(\bar{f} , P') – U(f,P') \leq 2. D.\frac{\epsilon}{4D} = \epsilon/2 $

now,

$ = U(f)$

$ > U(f , P_{0}) – \epsilon/2$ because we choose $P_0$ that way

$ \geq U(f,P') – \epsilon/2$ cause $P'$ is refinement of $P_{0}$

$ \geq U(\bar{f},P') – \epsilon/2 – \epsilon/2$

$ = U(\bar{f} , P') – \epsilon$

since this holds for arbitrary $\epsilon$ we have that $U(f) = \inf\{U(\bar{f},P) : P \in \mathcal{P}\}$ similarly we can show that $L(f) = \sup\{L(\bar{f},P) : P \in \mathcal{P}\} $ and we are done.

Best Answer

I think your proof is fine.

Alternately, note that you could make things very simple :

  • Prove that the functions $\{1_{\{c\}} : c \in [a,b]\}$ where $1_{\{c\}}$ is the indicator function of $\{c\}$, are Riemann integrable and have Riemann integral zero. This is much easier to do, because you can choose your partitions and compute the upper and lower sum obviously.

  • From here, obviously by scaling, functions of the form $L 1_{\{c\}}$ are integrable for any number $L$, and their integral is zero.

  • Suppose $\bar f$ differs from $f$ at finitely many points. Then $\bar f - f$ is a finite sum $\sum_{i=1}^N L_i1_{x_i}$, which is Riemann integrable with integral zero. It follows that $\bar f$ is integrable with the same integral as $f$.

EDIT : That the Riemann integral of $1_{\{c\}}$ over $[a,b]$, for $c \in [a,b]$ is zero, follows by considering the partition $P_{\epsilon} : a<c-\epsilon<c+\epsilon<b$ of mesh size $\epsilon$ (for $\epsilon>0$ small enough), which is such that $U(1_{\{c\}},P_{\epsilon}) =2\epsilon$ and $L(f,P_{\epsilon}) = 0$. Therefore, by taking $\epsilon \to 0$, we must have $U(f) \leq 0$, while $L(f) \geq 0$. Hence, $U(f) = L(f)$, as desired.

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