91Ó°ÊÓ

Weak convergence is a key concept in measure theory, which helps us understand how sequences of probability measures approach a limiting measure. In our case, we are given a sequence of measures \( \mu_n = \frac{1}{n} \sum_{k=0}^{n} \delta_{k/n} \) and need to show that it converges weakly to the Lebesgue measure \( \mu \) on the interval \([0, 1]\).

A measure \( \mu_n \) is said to converge weakly to \( \mu \) if, for every bounded and continuous function \( f \), the integrals of \( f \) with respect to \( \mu_n \) tend towards the integral of \( f \) with respect to \( \mu \). Formally, this can be expressed as:

• \( \int f \, d\mu_n \rightarrow \int f \, d\mu \)

The beautiful part about weak convergence is that it doesn't require the measures to converge pointwise, making it a powerful tool in analysis and probability theory. In practical terms, it shows how the distribution of values represented by \( \mu_n \) behave like those of the limiting measure \( \mu \), which in this problem is the Lebesgue measure restricted to \([0,1]\).

The Lebesgue measure, \( \lambda \), is fundamental in measure theory, particularly in providing a rigorous basis for integrals beyond the simple Riemann integral.

In the given exercise, the Lebesgue measure is restricted to the interval \([0,1]\), allowing us to define \( \mu \) as a measure that essentially captures the length of intervals within \([0,1]\). Unlike discrete measures that attach a mass to single points, the Lebesgue measure evaluates the 'size' of a set in a continuum, which is pivotal for defining integrals of functions over intervals.

The Lebesgue measure is more powerful and flexible than the Riemann integral for several reasons:

• It can handle a wider class of functions, including those with discontinuities or on more complex sets.
• It allows for the use of limits, which are essential in understanding weak convergence.

In the exercise, we demonstrate weak convergence using the Lebesgue measure since it enables the approximation of the integral of any bounded continuous function \( f \) over \([0, 1]\). This approximation aligns with how the Riemann sums approximate that same integral as \( n \) grows.

Riemann sums are a classical method used to approximate the integral of a function over an interval. They serve as the bridge between discrete approximations and continuous definitions.

In the provided exercise, each measure \( \mu_n \) corresponds to a sum of point masses situated at equally spaced intervals \( k/n \), weighted by \( 1/n \). These sums can be visualized as partitions of the interval \([0,1]\), where each partition contributes to approximating the integral.

The Riemann sum defined by \( \frac{1}{n} \sum_{k=0}^{n} f(k/n) \) closely resembles our conventional understanding of how to approximate \( \int_0^1 f(x) \, dx \).

• The partition of \([0,1]\) becomes finer as \( n \) increases, making the approximation more accurate.
• This approach ensures that \( \int f \, d\mu_n \rightarrow \int_0^1 f(x) \, dx \) as \( n \rightarrow \infty \).

By demonstrating that these Riemann sums converge to the integral, we affirm that \( \mu_n \) converges weakly to the Lebesgue measure \( \mu \). This not only illustrates the utility of Riemann sums in approximating integrals but also shows how weak convergence ties abstract concepts within measure theory to tangible mathematical tools.