Could you explain how to do this with at most detail? **Problem Statement**Let \( P \) be the Cantor set. Let \( f \) be a bounded real-valued function on \([0,1]\) which is continuous at every point outside \( P \). Using the definition of Riemann integration, prove that \( f \) is Riemann integrable. **Discussion**The problem involves proving the Riemann integrability of a function \( f \) defined on the interval \([0,1]\). The key condition here is that the function is continuous everywhere except possibly on the Cantor set \( P \). The Cantor set \( P \) is known for being a set of measure zero, which plays a crucial role in determining the integrability of \( f \). By the property of Riemann integration, a bounded function on a closed interval is Riemann integrable if the set of discontinuities has measure zero. Hence, leveraging these properties should guide the proof.

Given that P be the cantor set. Let f be a bounded real valued function on 0,1, which is continuous at every point outside P. We know that he cantor set can be obtained iteratively by removing 13 open subintervals of 0,1. Suppose at nth iteration Pn be the remaining portion of 0,1. Therefore at nth iteration P=∪Pn and length of P is 23n, which is converges to 0. Therefore we can enclosed these Pn by open intervals with negligible larger length. Since the given function f is bounded, therefore there exists M≥0 such that M=supfx: x∈0,1. Suppose ε>0 be an arbitrary positive real number. Let ∪i=1nai,bi be collection of open interval which covers the cantor set P such that ∑i=1nbi-ai<ε.Since ∪i=1nai,bi is open set, therefore K=0,1-∪i=1nai,bi is a compact set. Therefore the given function is uniformly continuous on K. Hence there exists δ>0 such that fx-fy<ε if x-y<δ and x,y∈K. Let Q=x0,x1,…,xr be a partition of 0,1, which includes the endpoints of collection of open interval ∪i=1nai,bi and ∆xi<δ. That is ai,bi∈Q. let mi=inffx:x∈xi-1,xi and Mi=supfx:x∈xi-1,xi, where i=1,2,…,r. Observe that Mi-mi≤2M for all i=1,2,…,r. Also observe that Mi-mi<ε, whenever xi-1,xi∈K. Therefore: UQ,f-LQ,f=∑iMi-mi∆xi=∑xi-1,xi∉KMi-mi∆xi+∑xi-1,xi∈KMi-mi∆xi≤∑xi-1,xi∉K2M∆xi+∑xi-1,xi∈Kε∆xi<2Mε+ε1-0=2M+1ε If we take ε1=ε2M+1, then UQ,f-LQ,f<ε1. Hence the given function is Riemann integrable on 0,1.