The interval [a,a] has no length and has Lebesgue measure zero, but it is not the null set .... I'm missing something fundamental, I think. Given a measurable space $(\Omega, \mathcal{F})$ we define a measure $\mu: \mathcal{F} \rightarrow \mathbb{R}$ as having two properties, the first of which, $\mu(\emptyset) = 0$, is my problem.
The Lebesgue measure of intervals of the form $(a,b)$ is $\mathcal{l}((a,b)) = b-a.$ If b=a the value of the measure is $0$ but the set is not the null set. So, how do we know $\mathcal{l}$() satisfies the first condition and is therefore a measure? Do we just define it to be a property? If yes then all is good. If no, then I am missing something fundamental. I fear the answer is no.
 A: This answer summarizes my comments above.
It was assumed that $L$ is a function that maps intervals of the type $(a,b)$ to the length of the interval:
$$L((a,b)) = b-a \quad \forall a,b \in \mathbb{R}, a<b$$
This function $L$ is currently only defined on subsets of $\mathbb{R}$ that are intervals of the form $(a,b)$ where $a, b$ are real numbers and $a<b$.  This collection of intervals does not form a sigma algebra on $\mathbb{R}$. Also, this collection of intervals does not include important subsets of $\mathbb{R}$ such as
\begin{align}
&\phi\\
&\{3.5\}\\
&[3.5, 5)\\
&[0,1] \cup [3.5, 5)\\
&\{x \in [0,1] : x \notin \mathbb{Q} \}
\end{align}
However, it can be shown that the sigma algebra generated by the intervals of the type $(a,b)$ is equal to $\mathcal{B}(\mathbb{R})$ (the standard Borel sigma algebra on $\mathbb{R}$). The collection of sets $\mathcal{B}(\mathbb{R})$ includes all intervals of the type $(a,b)$, but also includes many more subsets of $\mathbb{R}$ (including the ones listed above). An important question is whether or not we can define $L(A)$ on all of the additional sets $A \in \mathcal{B}(\mathbb{R})$ that are not of the form $(a,b)$ in such a way that the extended function $L:\mathbb{B}(\mathbb{R})\rightarrow [0,\infty]$ is a valid measure, and in a way that preserves the value $L((a,b))=b-a$ for all of the sets $(a,b)$ on which $L$ was originally defined.
A deep theorem of measure theory called the Caratheodory extension theorem proves that such an extension of $L$ exists and is unique. So we must define $L(A)$ for the additional sets $A$, but there is one and only one way to define these values while being consistent with the requirements of a measure. This is like a Sudoku puzzle where some values of $L(A)$ are already given to us, we must find the missing values $L(A)$ for the remaining sets $A$, and there is only one way to correctly do it.
We know that such an extension requires us to define $L(\phi)=0$ simply because all measures must satisfy that. Also, assuming $L$ is the extended measure, we see that for all $x\in \mathbb{R}$ and all $\epsilon>0$
$$ \phi \subseteq \{x\} \subseteq (x-\epsilon, x+\epsilon)$$
and so we must have
$$ 0 \leq L(\{x\})\leq L((x-\epsilon, x+\epsilon)) = 2\epsilon$$
This holds for all $\epsilon>0$, and so we must have $L(\{x\})=0$ for all $x \in \mathbb{R}$. From this, together with basic properties of a measure, we can also infer
\begin{align}
&L([3.5, 5)) = L(\{3.5\}\cup (3.5,5)) = L(\{3.5\})+L((3.5, 5)) = 0 + 1.5 = 1.5\\
&L([0,1])=1\\
&L([0,1]\cup [3.5,5)) = 2.5\\
&L([0,1]\cap \mathbb{Q}) = L\left(\cup_{x \in [0,1]\cap \mathbb{Q}}\{x\} \right) = \sum_{x \in [0,1]\cap \mathbb{Q}} L(\{x\})=0\\
&L(\{x \in [0,1]: x \notin \mathbb{Q}\}) = 1
\end{align}
