contradiction to the condition
Not what you're looking for?
Let (X, E, u) be a measure space with u a non-negative
measure. Suppose that
1. f : X -> R is measurable
2. f (x) >= 0 a.e. with respect to u.
3. integral (over X) f du = 0
Prove that f (x) = 0 a.e. with respect to u.
note: E = Capital Sigma
u = lowercase mu
Purchase this Solution
Solution Summary
A contradiction to the condition is embedded.
Solution Preview
Proof:
is a measurable space with a non-negative measure.
First, a.e., with respect to . It means that except a ...
Purchase this Solution
Free BrainMass Quizzes
Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
Probability Quiz
Some questions on probability
Solving quadratic inequalities
This quiz test you on how well you are familiar with solving quadratic inequalities.
Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts