Answer :
Let's carefully go through each statement one by one, ensuring we understand their validity in the context of bounded functions and partitions on [tex]\([a, b]\)[/tex].
### (i) [tex]\(L(P \cup Q, f) \leq L(P, f)\)[/tex]
Explanation:
- The lower sum [tex]\(L(P, f)\)[/tex] for a partition [tex]\(P\)[/tex] is the sum of areas of rectangles under the curve, where the height of each rectangle is determined by the infimum of [tex]\(f\)[/tex] on each subinterval.
- When we combine [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] into [tex]\(P \cup Q\)[/tex], the partition [tex]\(P \cup Q\)[/tex] may have more points than [tex]\(P\)[/tex] alone.
- Having more partition points generally means more subintervals or potentially smaller subintervals. A finer partition can only make the approximation of the area more accurate or better.
- Hence, this refined partition cannot lead to a lower sum greater than the original lower sum because we are taking the infimum, and more points give us more opportunities to find smaller values.
Thus, combining more points (i.e., [tex]\(P \cup Q\)[/tex]) into the lower sum cannot result in a sum larger than the lower sum based on fewer points. Therefore, [tex]\( L(P \cup Q, f) \leq L(P, f) \)[/tex].
### (ii) [tex]\(U(P \cap Q, f) \leq U(Q, f)\)[/tex]
Explanation:
- The upper sum [tex]\(U(Q, f)\)[/tex] for a partition [tex]\(Q\)[/tex] is the sum of areas of rectangles above the curve, where the height of each rectangle is determined by the supremum of [tex]\(f\)[/tex] on each subinterval.
- Similarly, when we consider [tex]\(P \cap Q\)[/tex], the partition [tex]\(P \cap Q\)[/tex] will have fewer points than [tex]\(Q\)[/tex] if [tex]\(Q\)[/tex] has any points not present in [tex]\(P\)[/tex], effectively taking the minimum of the points in both partitions.
- A finer partition will yield a more precise or smaller upper sum because the supremum over smaller intervals will be smaller or cannot increase.
- Thus, taking the intersection of partitions that might reduce the subintervals’ size won't make the upper sum larger.
So, the upper sum over the intersection [tex]\(P \cap Q\)[/tex] cannot exceed the upper sum over just [tex]\(Q\)[/tex]. Hence, [tex]\( U(P \cap Q, f) \leq U(Q, f) \)[/tex].
### (iii) [tex]\(L(P, f) \leq U(P \cup Q, f)\)[/tex]
Explanation:
- [tex]\(L(P, f)\)[/tex] is a lower sum, and [tex]\(U(P \cup Q, f)\)[/tex] is an upper sum.
- Generally, for the same partition, the upper sum is always at least as large as the lower sum because the supremum will be greater than or equal to the infimum over any subinterval.
- When we refine the partitions by taking [tex]\(P \cup Q\)[/tex], the upper sum [tex]\(U(P \cup Q, f)\)[/tex] represents an even finer approximation, which could only make the upper sum more accurate without reducing its value below the lower sum.
Therefore, for any such partitions, the lower sum will not exceed the upper sum of [tex]\(P \cup Q\)[/tex], hence [tex]\( L(P, f) \leq U(P \cup Q, f) \)[/tex].
### (iv) [tex]\(U(P, f) \leq U(P \cup Q, f)\)[/tex]
Explanation:
- Similar to the consideration about the lower sums, the upper sum with more points (i.e., [tex]\(P \cup Q\)[/tex]) will generally capture the supremum more accurately or with smaller intervals.
- Including more partition points can only keep the upper sum the same or make it more precise (often making it smaller in larger intervals). However, it can never increase the upper sum.
Thus, [tex]\( U(P, f) \leq U(P \cup Q, f) \)[/tex].
To summarize:
[tex]\[ (i) \quad L(P \cup Q, f) \leq L(P, f) \\ (ii) \quad U(P \cap Q, f) \leq U(Q, f) \\ (iii) \quad L(P, f) \leq U(P \cup Q, f) \\ (iv) \quad U(P, f) \leq U(P \cup Q, f) \][/tex]
### (i) [tex]\(L(P \cup Q, f) \leq L(P, f)\)[/tex]
Explanation:
- The lower sum [tex]\(L(P, f)\)[/tex] for a partition [tex]\(P\)[/tex] is the sum of areas of rectangles under the curve, where the height of each rectangle is determined by the infimum of [tex]\(f\)[/tex] on each subinterval.
- When we combine [tex]\(P\)[/tex] and [tex]\(Q\)[/tex] into [tex]\(P \cup Q\)[/tex], the partition [tex]\(P \cup Q\)[/tex] may have more points than [tex]\(P\)[/tex] alone.
- Having more partition points generally means more subintervals or potentially smaller subintervals. A finer partition can only make the approximation of the area more accurate or better.
- Hence, this refined partition cannot lead to a lower sum greater than the original lower sum because we are taking the infimum, and more points give us more opportunities to find smaller values.
Thus, combining more points (i.e., [tex]\(P \cup Q\)[/tex]) into the lower sum cannot result in a sum larger than the lower sum based on fewer points. Therefore, [tex]\( L(P \cup Q, f) \leq L(P, f) \)[/tex].
### (ii) [tex]\(U(P \cap Q, f) \leq U(Q, f)\)[/tex]
Explanation:
- The upper sum [tex]\(U(Q, f)\)[/tex] for a partition [tex]\(Q\)[/tex] is the sum of areas of rectangles above the curve, where the height of each rectangle is determined by the supremum of [tex]\(f\)[/tex] on each subinterval.
- Similarly, when we consider [tex]\(P \cap Q\)[/tex], the partition [tex]\(P \cap Q\)[/tex] will have fewer points than [tex]\(Q\)[/tex] if [tex]\(Q\)[/tex] has any points not present in [tex]\(P\)[/tex], effectively taking the minimum of the points in both partitions.
- A finer partition will yield a more precise or smaller upper sum because the supremum over smaller intervals will be smaller or cannot increase.
- Thus, taking the intersection of partitions that might reduce the subintervals’ size won't make the upper sum larger.
So, the upper sum over the intersection [tex]\(P \cap Q\)[/tex] cannot exceed the upper sum over just [tex]\(Q\)[/tex]. Hence, [tex]\( U(P \cap Q, f) \leq U(Q, f) \)[/tex].
### (iii) [tex]\(L(P, f) \leq U(P \cup Q, f)\)[/tex]
Explanation:
- [tex]\(L(P, f)\)[/tex] is a lower sum, and [tex]\(U(P \cup Q, f)\)[/tex] is an upper sum.
- Generally, for the same partition, the upper sum is always at least as large as the lower sum because the supremum will be greater than or equal to the infimum over any subinterval.
- When we refine the partitions by taking [tex]\(P \cup Q\)[/tex], the upper sum [tex]\(U(P \cup Q, f)\)[/tex] represents an even finer approximation, which could only make the upper sum more accurate without reducing its value below the lower sum.
Therefore, for any such partitions, the lower sum will not exceed the upper sum of [tex]\(P \cup Q\)[/tex], hence [tex]\( L(P, f) \leq U(P \cup Q, f) \)[/tex].
### (iv) [tex]\(U(P, f) \leq U(P \cup Q, f)\)[/tex]
Explanation:
- Similar to the consideration about the lower sums, the upper sum with more points (i.e., [tex]\(P \cup Q\)[/tex]) will generally capture the supremum more accurately or with smaller intervals.
- Including more partition points can only keep the upper sum the same or make it more precise (often making it smaller in larger intervals). However, it can never increase the upper sum.
Thus, [tex]\( U(P, f) \leq U(P \cup Q, f) \)[/tex].
To summarize:
[tex]\[ (i) \quad L(P \cup Q, f) \leq L(P, f) \\ (ii) \quad U(P \cap Q, f) \leq U(Q, f) \\ (iii) \quad L(P, f) \leq U(P \cup Q, f) \\ (iv) \quad U(P, f) \leq U(P \cup Q, f) \][/tex]
