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Anyway, from your question, it has given me an idea.

[nU] = n-1 if and only if n-1 <= nU < n if and only if (n-1)/n < U < 1. And probability is 1 - n/n + 1/n = 1/n.

[nU] = n-2 if and only if n-2 <= nU < n-1 if and only if (n-2)/n < U < (n-1)/n. Probability is 1/n again.

Continue with n-3, n-4 etc. Is it correct?

[nU] = n-1 if and only if n-1 <= nU < n if and only if (n-1)/n < U < 1. And probability is 1 - n/n + 1/n = 1/n.

[nU] = n-2 if and only if n-2 <= nU < n-1 if and only if (n-2)/n < U < (n-1)/n. Probability is 1/n again.

Continue with n-3, n-4 etc. Is it correct?

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U_n = [nU]/n. Compute P(U_(n+1) > U_n) and P(U_(n+1) = U_n) for each n in N (natural number) to show that the convergence is not monotone.

I have no idea how to do it at all and I will show you my attempt and I won't be surprised if I get it totally on the wrong track.

So here is my attempt:

[(n+1)U]/(n+1) > [nU]/n when k/(n+1) <= U < k/n , where k is an integer from 1 to n

[(n+1)U]/(n+1) = [nU]/n when U < 1/(n+1) or when U=1

So P(U_(n+1) > U_n) = summation (k/n - k/(n+1)), k from 1 to n = 1/2

P(U_(n+1) = U_n) = 1/(n+1)

Assuming that my probabilities are correct (please let me know if I am wrong), how do I show that the convergence is not monotone?

So, consider a sample space \(\Omega =\left [ 0,1 \right ] \) with a Uniform probability measure

Let \( m\left ( n \right )\) be a sequence of intervals of the form \( \left [ \frac{i}{k},\frac{\left ( i+1 \right )}{k} \right ] \) for \( i=0,...,k-1 \) and where \( k\in \mathbb{N} \), where

\( m\left ( 1 \right )=\left [ 0,1 \right ], \),\( m\left ( 2 \right )=\left [ 0,\frac{1}{2} \right ] \),..., \( m\left ( 6 \right )=\left [ \frac{2}{3},1 \right ] \).

Next, define a sequence of random variables on this sample space as \( X_{n}\left ( \omega \right )=\delta \left \{ \omega ;m\left ( n \right ) \right \} \) for

\( \omega \in \left [ 0,1 \right ] \).

Let \( \varepsilon > 0 \) and note that \( X_{n}\left ( \omega \right ) \) is 1 only on the interval \( m\left ( n \right ) \) such that

\( P\left ( \left |X _{n} \right |< \varepsilon \right )\leq 1-\frac{k}{n} \),

where \( k\left ( n \right ) \) is a sequence that converges to 0, \( n\rightarrow \infty \).

It follows that for any \( \varepsilon > 0 \),

\( \lim_{n\rightarrow \infty }P\left ( \left |X _{n} \right |<\varepsilon \right )=1 \),

so then it necessarily follows that \( X_{n}\overset{p}{\rightarrow}0 \) as \( n \to \infty \).

So, consider a sample space \(\Omega =\left [ 0,1 \right ] \) with a Uniform probability measure

Let \( m\left ( n \right )\) be a sequence of intervals of the form \( \left [ \frac{i}{k},\frac{\left ( i+1 \right )}{k} \right ] \) for \( i=0,...,k-1 \) and where \( k\in \mathbb{N} \), where

\( m\left ( 1 \right )=\left [ 0,1 \right ], \),\( m\left ( 2 \right )=\left [ 0,\frac{1}{2} \right ] \),..., \( m\left ( 6 \right )=\left [ \frac{2}{3},1 \right ] \).

Next, define a sequence of random variables on this sample space as \( X_{n}\left ( \omega \right )=\delta \left \{ \omega ;m\left ( n \right ) \right \} \) for

\( \omega \in \left [ 0,1 \right ] \).

Let \( \varepsilon > 0 \) and note that \( X_{n}\left ( \omega \right ) \) is 1 only on the interval \( m\left ( n \right ) \) such that

\( P\left ( \left |X _{n} \right |< \varepsilon \right )\leq 1-\frac{k}{n} \),

where \( k\left ( n \right ) \) is a sequence that converges to 0, \( n\rightarrow \infty \).

It follows that for any \( \varepsilon > 0 \),

\( \lim_{n\rightarrow \infty }P\left ( \left |X _{n} \right |<\varepsilon \right )=1 \),

so then it necessarily follows that \( X_{n}\overset{p}{\rightarrow}0 \) as \( n \to \infty \).

What I am showing is convergence in probability. It will not converge almost surely. As I said, convergence in probability is weaker than convergence almost surely.

In terms of whether your question regarding monotonicity: Are you being asked to use the Lebesque's Monotone Convergence Theorem???...I guess, I'm just not sure why you asking this.

Well, the problem asked me to find those probabilities and to use them to show that the convergence is not monotone.

m(1)=[0, 1], m(2)=[0, 1/2], m(3)=[1/2, 1], m(4)=[0, 1/3], m(5)=(1/3, 2/3), m(6)=[2/3, 1], m(7)=[0, 1/4], m(8)=[1/4, 1/2], m(9)=[1/2, 3,4], m(10)=[3/4, 1].