lebesgue integral

\[ 0 \leq g_n \uparrow g \Rightarrow \int g_n \, d\mu \uparrow \int g \, d\mu \] and \[ 0 \leq g_n \uparrow g, \text{a.e.} \Rightarrow \int g_n \, d\mu \uparrow \int g \, d\mu \] This says that if the sequence \(\{g_n\}\) pointwise converges to \(g\), then the Lebesgue integral of \(g\) exists and is equal to the pointwise limit the integrals of \(\{g_n\}\). This is noteworthy, because the same cannot be said of the Riemann integral (see below).

\[ \int(g + h)\, d\mu = \int g \, d\mu + \int h \, d\mu \] When \(\mu\) is a probability measure and \(g\) and \(h\) are random variables (see below), then we get the linearity of expectation (see properties of expectation).

If \((\Omega, \mathcal{F}, \mathbb{P})\) is a probability space and \(X:\Omega \rightarrow \bar{\mathbb{R}}\) is measurable (i.e. a random variable), then \(\int X \, d\mathbb{P}\) is also denoted \(\mathbb{E}[X]\) and called the expectation of \(X\).

Recall the definition of the Riemann integral \(\int_{a}^{b} f(x) \, dx\). We divide the interval \([a,b]\) using a finite sequence of points \(\sigma = (x_1, ..., x_n)\) that satisfy \(a = x_1 < x_2 \cdots < x_n = b\). Then, we define: \[ U(\sigma) = \sum_{i=1}^{n-1}\left(\max_{x_i \leq x < x_{i+1}} g(x)\right) \cdot (x_{i+1} - x_i) \] and \[ L(\sigma) = \sum_{i=1}^{n-1}\left(\min_{x_i \leq x < x_{i+1}} g(x)\right) \cdot (x_{i+1} - x_i) \]

\(U(\sigma)\) and \(L(\sigma)\) are approximations of the area under \(g\) from above and below respectively. You can see that they are the sums of rectangles, where the rectangles in \(U\) have heights above \(g\) and the rectangles in \(L\) have heights below \(g\).

Then, given a sequence of \(\sigma\) 's, the Riemann integral of \(g\) is defined as: \[ \lim\sup_{\sigma} L(\sigma) = \lim\inf_{\sigma} U(\sigma) = c \] provided that the above two quantities exist and are equal.

Let \(1_{\mathbb{Q}}\) be the Dirichlet function: \[1_{\mathbb{Q}} = \begin{cases} 1 & \text{if } x\in \mathbb{Q}\\ 0 & \text{if } x\not\in \mathbb{Q} \end{cases}\]

Then, for every interval \([x_i, x_{i+1})\), there is a rational number and a irrational number. So, we see that \(L(\sigma) = 0\) and \(U(\sigma) = 1\) for any \(\sigma\). So the integral does not exist.

From the wikipedia page.

Let's define a sequence of functions, such that each function is Riemann integrable, but the limit is not Riemann integrable.

Let \(\{a_k\}\) be a countable enumeration of all rational numbers.

Define function \(g_k\) where \[g_k(x) = \begin{cases} 1 & \text{if } x = a_j, j\leq k\\ 0 & \text{otherwise} \end{cases}\] Then we see that \(g_k\) is 0 except at a finite number of points, so each \(g_k\) is Riemann integrable. But \(g_k \rightarrow 1_{\mathbb{Q}}\) when \(k\rightarrow \infty\), which is not integrable.

A function \(g : \Omega \rightarrow \mathbb{R}\) is simple if it is Lebesgue measurable, finite, and can only take finitely many values. In particular, a simple function can be written as: \[\begin{equation} g(\omega) = \sum_{i=1}^{k} a_i 1_{A_i}(\omega) \label{eq1} \end{equation}\]

for all \(\omega \in \Omega\). That is, it can be written as a weighted sum of indicator functions – think of a generalized piecewise linear function.

Note that the representation on the LHS is not necessarily unique for \(g\). However, it is if we require the \(a_i\) to be unique and the sets \(A_i\) to form a partition of \(\Omega\). If this is true, then it is called a canonical representation.

Then, for simple functions \(g\) of the form in \(\ref{eq1}\), we define the Lebesgue integral: \[ \int g \, d\mu = \sum_{i=1}^{k} a_i \mu(A_i) \] where if \(a_i = 0\) and \(\mu(A_i) = \infty\), we assume \(a_i\mu(A_i) = 0\)

Note that this does not require \(g\) to be in canonical form. Quick check: if \(g\) has two distinct forms, will the integral for these forms be equal? Yes, because both forms will be equivalent to the canonical form, which implies that both forms have the same integral of the canonical form.

For a non-negative (not necessarily) simple function \(g: \Omega \rightarrow [0, \infty]\), let \(S(g)\) be the set of all non-negative simple functions \(q\) that satisfy \(0\leq q \leq g\). Then, we define the Lebesgue integral for \(g\) as: \[ \int g \, d\mu = \sup_{q\in S(g)} \int q \, d\mu \]

The above definition tells us how to find \(\int g \, d\mu\) given a \(S(g)\), but doesn't tell us how to find \(S(g)\). Here, we explicity give one example to help with understanding. For a measurable function \(g\), we will define a sequence \(q_n\) such that \(\int q_n \, d\mu \rightarrow \int g \, d\mu\).

For any positive integer \(r\), we define \(q_r : \Omega \rightarrow \mathbb{R}\): \[q_r(\omega) = \begin{cases} r &\text{if} g(\omega) \geq r\\ \frac{1}{2^r} &\text{if } \frac{i}{2^r} \geq g(\omega) \frac{i+1}{2^r}, i=0, 1, ...,r2^r - 1 \end{cases}\]

In words: \(q_r\) is capped by \(r\), otherwise it snaps to the nearest multiple of \(\frac{1}{2^r}\), rounding down. The resolution gets better as \(r\) increases.

Then, the two things to notice are:

• every \(q_r\) is simple and measurable.
  • Simple because the range is finite and contains finitely many values
  • Measurable because \(g\) is measurable, so any interval \(\frac{i}{2^r} \leq g(\omega) \leq \frac{i+1}{2^r}\) has a measurable pre-image.
• We have \(q_r \uparrow g\). That is for every \(q_r(\omega) \uparrow g(\omega)\) for every \(\omega\)

Now, you can use the mental image of these functions to understand what is going on in \(S(g)\).