# Haar & Lebesgue Measure

Basics of Probability

Our expectations and integrals are examples of Lebesgue integrals. How do these relate to the Riemann integrals we learned about in calculus?

The group of real numbers under addition $(\mathbb R, +)$ is an example of a locally compact group. Any Locally compact group $(G,+)$ can be topologized and the $\sigma$-algebra generated by the $\pi$-system of open sets will be called the Borel $\sigma$-algebra and denoted $\mathcal B(G)$. This is consistent with our notation for $\mathbb R$ and $\mathbb R^N$, both of which are locally compact abelian groups.

Lebesgue measure $\lambda$ on $\mathcal B(\mathbb R)$ is the unique measure for which $\lambda(a, b) = b – a$ for all open intervals $(a, b)$. It can be easily seen that $\lambda$ is also translation invariant: If $A \in \mathcal B(\mathbb R)$ and $x \in \mathbb R$, then $\lambda(x + A) = \lambda(A)$. It turns out that translation invariance is almost enough to completely characterize $\lambda$. Indeed, the only translation invariant measure for which $\lambda(0,1) = 1$ is Lebesgue measure.

This can be generalized to other locally compact abelian groups. Suppose $(G, +)$ with Borel $\sigma$-algebra $\mathcal B(G)$, and suppose $K \in \mathcal B(G)$ is a compact set containing an open set. Then there is a unique translation invariant measure $\lambda_G$ such that $\lambda_G(K) = 1$. This measure is called the Haar measure on $G$ normalized on $K$. When $G$ is in fact compact itself, it is natural to take $K=G$ in which case $\lambda_G$ becomes a probability measure on $(G, \mathcal B(G))$.

Exercise: Show that Haar measure on $\mathbb R^N$ is equal to the product measure $\lambda^N$ of Lebesgue measure $\lambda$ on $\mathbb R$.

#### Lebesgue Measure and Riemann Integrals

Suppose $f$ is a Riemann integrable function supported on the interval $(a, b]$. Then, given a partition $a = a_0 < a_1 < \ldots < a_N = b$, we may make the lower Riemann approximation to $f$ by setting $$f_{\mathbf a} = \sum_{n=1}^N \inf\{f(y) : y \in (a_{n-1}, a_n]\} \boldsymbol 1_{(a_{n-1}, a_n]}.$$ The associated Riemann sum is then $$R_{\mathbf a}(f) = \sum_{n=1}^N \inf\{f(y) : y \in (a_{n-1}, a_n]\} (a_n – a_{n-1}).$$

If $(\mathbf a_n)$ is a sequence of refined partitions, such that $$R_{\mathbf a_n}(f) \rightarrow \int_a^b f(x) \, dx$$ then $f_{\mathbf a_n}$ is a sequence of measurable simple functions which increase to $f$, and $$\int f d\lambda = \lim \int f_{\mathbf a_n} d\lambda = \lim R_{\mathbf a_n}(f) = \int_a^b f(x) \, dx.$$ We have essentially proved the following theorem.

Theorem: If $f$ is a Riemann integrable function on $\mathbb R$, then $f$ is Borel measurable and $$\int f d\lambda = \int_{\mathbb R} f(x) \, dx.$$ Moreover, if $g : \mathbb R^M \rightarrow \mathbb R$ is Riemann integrable, then $g$ is Borel measurable and $$\int g \, d\lambda^M = \int_{\mathbb R} \cdots \int_{\mathbb R} g(x_1, \ldots, x_M) \, dx_1 \cdots dx_M.$$