Written by Chandra Gummaluru
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What is Lebesgue Integration?

Integration can be used to find areas, volumes, central points, and many useful things. Typically, it is introduced as a tool that can be used to find the area under a curve defined by a function.

The Reimann Integral and its Problems

Suppose we have a function, $f: \mathbb{R} \rightarrow \mathbb{R}$ for which we seek the area under its graph within an interval, $[a,b]$. We will denote this quantity as $\int_{a}^{b}f(x)dx$.

To do this, we pick $n$ points $x_1, \dots, x_n \in [a,b]$, so that $a = x_1 < x_2 < \dots < x_{n-1} < x_{n} = b$. The expression, \[ \sum_{i=1}^{n-1}(x_{i+1}-x_i)\inf_{x \in [x_i,x_{i+1}]}f(x) \tag{1a}\label{eq_reimann_lower} \] is called the lower-sum, and it under-approximates the desired area. Similarly, the expression \[ \sum_{i=1}^{n-1}(x_{i+1}-x_i)\sup_{x \in [x_i,x_{i+1}]}f(x) \tag{1b}\label{eq_reimann_upper} \] is called the upper-sum and it over-estimate the desired area. As $n$​ grows large, both of these approximations get better. Thus, if we can show that both approximations approach the same value in the limit the area can be defined as follows: \[ \int_{a}^{b}f(x)dx := \sum_{i=1}^{n-1}(x_{i+1}-x_i)\inf_{x \in [x_i,x_{i+1}]}f(x) = \sum_{i=1}^{n-1}(x_{i+1}-x_i)\sup_{x \in [x_i,x_{i+1}]}f(x) \tag{1c}\label{eq_reimann_integral} \]

This is called the Reimann integral. The problem is that both approximations need not always approach the same value even if the area under seems intuitive. For example, consider a function, $I_{\mathbb{Q}}$, defined so that \[ I_{\mathbb{Q}}(x) = \begin{cases} 1, x \in \mathbb{Q}\\0, \text{ otherwise} \end{cases} \] Intuitively, we expect the area to be zero since $\mathbb{Q}$ is countably infinite. Every sub-interval of $[a,b]$, where $a \neq b$ contains at least one irrational number, and so, the lower-sum is indeed always zero. However, since $\mathbb{Q}$ is dense, any sub-interval of $[a,b]$ will contain at least one rational number, and so, the upper-sum is always one. Thus, the Reimann integral does not exist.

Generalizing Integration

One issue with the Reimann integral is that it is difficult to generalize for a function with an arbitrary domain since it is not clear what an appropriate partitioning of the domain would be or how to measure the size of the resulting partitions. Our goal is to define a more general type of integral.

We consider a function, $f: \Omega \rightarrow \mathbb{R}$, where $\Omega$ is a non-empty but arbitrary set. If $f$ is bounded, we can pick $n$ points in its range so that $\min_{\Omega}f = y_1 < \dots < y_n = \max_{\Omega}f$. The expression, \[ \sum_{i=1}^{n-1}y_i\mu(f^{-1}([y_i,y_{i+1}])) \tag{2a}\label{eq_lebesgue_lower} \] under-estimates the area while the expression \[ \sum_{i=2}^{n}y_i\mu(f^{-1}([y_{i-1},y_{i}])) \tag{2b}\label{eq_lebesgue_upper} \] where $f^{-1}([a,b])$ denotes the inverse image of $[a,b]$ under $f$, and $\mu: \Omega \rightarrow \mathbb{R}$ measures the size of subsets of $\Omega$.

Measuring Sets

We need to understand the properties of $\mu$. Let $\Sigma$ denote the set of subsets of $\Omega$ that we need to measure. We will refer to such subsets as $\mu$-measurable subsets. We expect a few properties to hold:

Intuitively, these properties mean ensure that combining measurable sets results in a measurable set. If $\Sigma$ satisfies these properties, we call it a $\sigma$-algebra. Obviously (P1.1) and (P1.2) imply that $\emptyset \in \Sigma$. It also turns out that (P1.2) and (P1.3) imply (P1.4). Thus, (P1.4) is often not explicitly stated.

There are many $\sigma$-algebras for any $\Omega$, and thus, many measure spaces. The largest one is $(\Omega, \mathcal{P}(\Omega))$, where $\mathcal{P}(\Omega)$ denotes the power-set of $\Omega$. Of course, we are often interested in the smallest one. It can be shown that the smallest $\sigma$-algebra of $\Omega$ that contains $M \subseteq \Omega$ is given by \[\sigma(\Omega, M) = \bigcap_{\substack{\Sigma \in \sigma(\Omega)\\M \subseteq \Sigma}}\Sigma,\] where $\sigma(\Omega)$ is the set of all $\sigma$-algebras of $\Omega$.

We can now define a measure, $\mu: \Omega \rightarrow \mathbb{R}$. Again, we expect a few intuitive properties to hold:

  1. the measure of the empty-set is zero, i.e., $\mu(\emptyset) = 0$
  2. the measure of the union of disjoint sets is the sum of the measures of the individual sets, i.e., \[\mu\left(\bigcup_{i}A_i\right) = \sum_{i}\mu(A_i), A_i \in \Sigma, i \in \mathbb{N}, A_i \cap A_j = \emptyset\]

In general, a mapping $f: \Omega \rightarrow \Omega’$ is measurable w.r.t. $\Sigma$ and $\Sigma’$ iff \[ f^{-1}(S) = \lbrace s \in \Omega \text{ s.t. } f(s) \in S \rbrace \in \Sigma, \forall S \in \Sigma’ \] We write $f: (\Omega, \Sigma) \rightarrow (\Omega’, \Sigma’)$ to denote that $f$ is measurable w.r.t. $\Sigma$ and $\Sigma’$.

Lebesgue Integration

Let $\Omega$ be a non-empty set and $\Sigma$ be a $\sigma$-algebra on $\Omega$. If $f: (\Omega, \Sigma) \rightarrow (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ is bounded, and $R \in \Sigma$ , then the lower and upper sums defined by (\ref{eq_lebesgue_lower}) and (\ref{eq_lebesgue_upper}) are not only well defined, but also converge to the same value in the limit as $n \rightarrow \infty$. Thus, we define the integral as: \[\int_{R}f(\omega)d\mu(\omega) := \lim_{n\rightarrow\infty}\sum_{i=1}^{n-1}f(y_i)\mu(f^{-1}([y_i,y_{i+1}])) = \lim_{n\rightarrow\infty}\sum_{i=2}^{n}f(y_i)\mu(f^{-1}([y_{i-1},y_{i}]))\tag{2c}\label{eq_lebesgue_integral}\]