Question:
If \(f,g\) are bounded functions defined on \([a,b]\) and let \(P\) be any partition of \([a,b]\), then which of the following is NOT TRUE?
If \(f,g\) are bounded functions defined on \([a,b]\) and let \(P\) be any partition of \([a,b]\), then which of the following is NOT TRUE?
Show Hint
For lower sums, \(L(P,f+g)\geq L(P,f)+L(P,g)\). For upper sums, \(U(P,f+g)\leq U(P,f)+U(P,g)\).
Updated On: Jul 3, 2026
- \(L(P,-f)=-U(P,f)\)
- \(U(P,-f)=-L(P,f)\)
- \(L(P,f+g)\leq L(P,f)+L(P,g)\)
- \(U(P,f+g)\leq U(P,f)+U(P,g)\)
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The Correct Option is C
Solution and Explanation
Concept:
For lower and upper sums, we use infimum and supremum on each subinterval. Important properties are: \[ L(P,-f)=-U(P,f) \] \[ U(P,-f)=-L(P,f) \] Also, \[ U(P,f+g)\leq U(P,f)+U(P,g) \] But for lower sums, \[ L(P,f+g)\geq L(P,f)+L(P,g) \]
Step 1: Analyze lower sum of \(f+g\). For any subinterval, \[ \inf(f+g)\geq \inf f+\inf g \] Therefore, after multiplying by the length of subintervals and summing, \[ L(P,f+g)\geq L(P,f)+L(P,g) \]
Step 2: Compare with option (C). Option (C) says \[ L(P,f+g)\leq L(P,f)+L(P,g) \] This is opposite to the correct inequality. Therefore, option (C) is not true.
Step 3: Final answer. \[ \boxed{L(P,f+g)\leq L(P,f)+L(P,g)} \]
For lower and upper sums, we use infimum and supremum on each subinterval. Important properties are: \[ L(P,-f)=-U(P,f) \] \[ U(P,-f)=-L(P,f) \] Also, \[ U(P,f+g)\leq U(P,f)+U(P,g) \] But for lower sums, \[ L(P,f+g)\geq L(P,f)+L(P,g) \]
Step 1: Analyze lower sum of \(f+g\). For any subinterval, \[ \inf(f+g)\geq \inf f+\inf g \] Therefore, after multiplying by the length of subintervals and summing, \[ L(P,f+g)\geq L(P,f)+L(P,g) \]
Step 2: Compare with option (C). Option (C) says \[ L(P,f+g)\leq L(P,f)+L(P,g) \] This is opposite to the correct inequality. Therefore, option (C) is not true.
Step 3: Final answer. \[ \boxed{L(P,f+g)\leq L(P,f)+L(P,g)} \]
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