Question

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.

Answer 1

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.