Bounding the cardinality of finite Index sets by the Lebesgue Measure I am trying to find an estimate for the cardinality of indices with the upper bound beeing the Lebesgue Measure and to generalize the statement.
Let $0 \leq a < b < \infty$ and $n \in \mathbb{N},$ then
$$ card\lbrace i \in \mathbb{N}: \, a \leq i/n \leq b \rbrace \leq n \cdot \lambda([a,b]) +1,$$
where $\lambda$ denotes the Lebesgue-Measure on $\mathbb{R}.$
Question 1: Are there situations, where the upper bound without the "+1" is correct?
Question 2: Let's say the set of interest changes from $\lbrace a \leq i/n \leq b \rbrace$ to an arbitrary (Lebesgue) measurable set $A \subset \mathbb{R}^+_0.$ Does the inequation
$$ card\lbrace i \in \mathbb{N}: \,i/n \in A \rbrace \leq n \cdot \lambda(A) +1$$
stil hold?
Question 3: Is it true and possible to show, that
$$ 1/n \cdot card\lbrace i \in \mathbb{N}: \,i/n \in A \rbrace 
\rightarrow_{n \to \infty} \lambda(A)$$
Riemann-Sums seem to be inapplicable in this case.
Many thanks and best wishes!
 A: This became a little too long to be just a comment.
Item 1:
a. How many intervals of length $\frac{1}{n}$ can fit completely in $[a,b]$ under the condition that any two sub-intervals have at most one extreme point as intersection?  Answer:
$\lfloor n(b-a) \rfloor$, where $\lfloor \cdot \rfloor$ is the floor function.
b. How many of the extremes points of such intervals will be $[a,b]$?  Answer:  $\lfloor n(b-a) \rfloor +1$
So, we would have $ card\lbrace i \in \mathbb{N}: \, a \leq i/n \leq b \rbrace = \lfloor n(b-a) \rfloor +1$.
In this case, we will never have $\lfloor n(b-a) \rfloor +1 \leq n(b-a)$,  , because
$n(b-a) - \lfloor n(b-a) \rfloor < 1$.
c. However, the sub-intervals don't need to begin at $a$. So, if
$$ \min \left \{\frac{i}{n}:  \frac{i}{n} \geq a \right\} -a + b - \max \left \{\frac{i}{n}: \frac{i}{n} \leq b \right\} \geq \frac{1}{n} $$
we have enough room so that the space before the first sub-interval plus the space after the last sub-interval be big enough for one sub-interval, but useless because those spaces are not connected to each other.
So, if
$$ \min \left \{\frac{i}{n}:  \frac{i}{n} \geq a \right\} -a + b - \max \left \{\frac{i}{n}: \frac{i}{n} \leq b \right\} \geq \frac{1}{n} $$
since
$$ \min \left \{\frac{i}{n}:  \frac{i}{n} \geq a \right\} -a <\frac{1}{n} $$
and
$$ b - \max \left \{\frac{i}{n}: \frac{i}{n} \leq b \right\} < \frac{1}{n} $$
we will have:
$$ card\lbrace i \in \mathbb{N}: \, a \leq i/n \leq b \rbrace = \lfloor n(b-a) \rfloor  \leq n(b-a)$$
Item 2: The answer is no. Just take $A=\Bbb Q$
and remember that $\lambda(\Bbb Q)=0$.
Item 3:  The answer is also no. Again take $A=\Bbb Q$. Then, for all $n \in \mathbb{N}$, $n>0$,
$ 1/n \cdot card\lbrace i \in \mathbb{N}: \,i/n \in A \rbrace = \infty $ and $\lambda(A)=0$. So
$$\lim_{n \to \infty}  1/n \cdot card\lbrace i \in \mathbb{N}: \,i/n \in A \rbrace = \infty \neq 0 = \lambda(A)$$
