The German Wikipedia claims that comes from "leer", which means "empty" in German. That seems plausible, as German used to be one of the major languages in mathematics.

Chomsky used as the empty string (or actually as the identity element for string concatenation) in his early papers. Some people in combinatorics still use as the empty string, with the same justification.

The empty string is a sequence of symbols from the alphabet with length zero. The empty string is never a symbol in the alphabet. Your language - the language of all strings over {0, 1} with no more than four 1s - includes the empty string, since the empty string contains fewer than four 1s. Therefore, your DFA must accept the empty string to accept the language. As you have already observed, to accept the empty string, the initial state must also be a final/accepting state.

We will play a game. There is a machine with buttons and a green lamp. Your job is to press the right sequence of buttons that make the green lamp light up.

For example, perhaps the machine has only one button, labeled a. When you press it, the lamp lights up! Then you press the button again and the lamp turns off. Then you press a third time and the lamp turns on again. You find that pressing the button always turns the lamp on if it was off, and off if it was on. To light up the lamp and win the game all you need to do is press the button an odd number of times. We say that the machine accepts the language a(aa)*, which is a regular expression that represents the set of all strings of as of odd length—all the possible winning sequences of button presses.

Now imagine an even simpler machine: the green lamp is already lit! And there is a sign over the button that says DON'T PRESS THE BUTTON. Indeed, if you do press the button the lamp goes off and no amount of pressing makes it turn on again. Don't you wish you had followed the sign's advice? This machine accepts only the string $\epsilon$, which is the string of zero button presses. The regular expression $\epsilon$ represents the set that contains this one string and nothing else.

Now imagine the simplest machine of all: the green lamp never lights up no matter what buttons you press. Here you can't win the game. The set of strings accepted by the machine is empty, which we write $\varnothing$.

The given grammar is: $$ S \to a \mid Ab \mid aBa \\ A \to b \mid \epsilon \\ B \to b \mid A $$

Now the algorithm for the removal of null productions is as follows:

1. We need to find all the nullable symbols from the grammar. A nullable symbol is one that generates the null string in one or more steps.

   Since $A \rightarrow \epsilon$ so the symbol $A$ is nullable.

   Due to $B\rightarrow A$ we shall have $B\Rightarrow ^* \epsilon$. So $B$ as well is nullable.

   These are the only two nullable symbols in the grammar.

2. Remove all null productions in the grammar, and use substitution for the nullable symbols to incorporate the effect which could have been obtained due to their nullable nature as follows:

   For each nullable symbol on the RHS of a production, rewrite the production, by once including it, and another time by not including it. Only with the exception that if due to this rewriting of rules, a null production is introduced, then just ignore that production.

   So for the given grammar, we have:

   For $S$ productions:

   $S \to a \mid Ab \mid b \mid aBa\mid aa$

   For the $A$ productions we have:

   $A\to b$

   For the $B$ productions we have:

   $B\to b$

Since $A$ and $B$ are unit productions, they can be removed by substituting their RHS in the productions for $S$.

So the final grammar:

$S \to a \mid bb \mid b \mid aba\mid aa$