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Let (X,F,λ) be the measure space where X=R,F=B(R) and λ is the Lebesgue measure.

A) If fₙ 1/n= χ[n,[infinity]). Prove that (fₙ ) is decreasing and converges uniformly to a function f, but that ∫fdλ ≠lim ₙ→[infinity] ∫fₙ dλ. This example shows that the Monotone Convergence Theorem is not valid for a sequence of decreasing functions.
B(R)=σ−algebra Borel
χ characteristic function.

User King Mort
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Final Answer:

A) The sequence of functions
\(f_n = \chi_([n, \infty))\) is decreasing and converges uniformly to a limit function f. However,
\(\int f d\lambda \\eq \lim_(n \to \infty) \int f_n d\lambda\), illustrating that the Monotone Convergence Theorem does not hold for decreasing functions.

Step-by-step explanation:

In this example, each function
\(f_n\) is defined as the characteristic function of the set
\([n, \infty)\), denoted by
\(\chi_([n, \infty))\). Firstly, we establish that
\(f_n\) is decreasing. As n increases, the set
\([n, \infty)\) becomes larger, causing
\(f_n\) to take the value 1 on a smaller set, hence decreasing.

Secondly, to show uniform convergence, for any
\(\varepsilon > 0\), we can choose N such that for all
\(n \geq N\),
\(\lVert f_n - f \rVert_\infty < \varepsilon\), where f is the pointwise limit of
\(f_n\). This implies that
\(f_n\) converges uniformly to f.

Finally, the failure of the Monotone Convergence Theorem is evident when evaluating the integrals. The integral of the limit function f is not equal to the limit of the integrals of
\(f_n\). This discrepancy arises due to the unbounded nature of the sets
\([n, \infty)\), leading to a failure in the conditions required for the Monotone Convergence Theorem.

User Shabby
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