Hint: Think about Bass's Proposition 4.14 (1): given any $\delta > 0$, there is an open set $G$ with $A \subset G$ and $m(G - A) < \delta$. Then recall (or prove) that every open set is a countable union of disjoint open intervals.
Here are some more details. An argument like yours has difficulties with the possibility $m(A) = \infty$, but we can reduce to the case $m(A) < \infty$ by intersecting $A$ with large bounded sets, as below.
Suppose $\epsilon \in (0,1)$ is such that $m(A \cap I) \le (1-\epsilon)m(I)$ for each interval $I$. Let $k$ be a positive integer and let $A_k = A \cap [-k,k]$. Note that $m(A_k) \le 2k < \infty$, and that for each interval $I$, we have $$m(A_k \cap I) \le m(A \cap I) \le (1-\epsilon) m(I). \tag{1}$$
Let $\alpha > 0$ be arbitrary, and using Proposition 4.14, choose an open set $G$ with $A_k \subset G$ and
$$m(G - A_k) < \alpha \epsilon. \tag{2}$$
In particular, $m(G) = m(A_k) + m(G-A_k) < 2k + \alpha \epsilon < \infty$.
Now $G$ can be written as a disjoint union of open intervals: $G = \bigcup_{n=1}^\infty I_n$. Note that $m(I_n) \le m(G) < \infty$ for each $n$. Also, $G - A_k = \bigcup_{n=1}^\infty (I_n - A_k)$ which is also a disjoint union.
Now since $I_n - A_k = I_n - (I_n \cap A_k)$, we have
$$m(I_n - A_k) = m(I_n) - m(I_n \cap A_k) \ge m(I_n) - (1-\epsilon)m(I_n) = \epsilon m(I_n) \tag{3}$$ using (1). So by countable additivity,
$$m(G - A_k) = \sum_{n=1}^\infty m(I_n - A_k) \ge \sum_{n=1}^\infty \epsilon m(I_n) = \epsilon m(G). \tag{4}$$
Combining (1) and (3), we get $\epsilon m(G) < \alpha \epsilon$, so $m(G) < \alpha$. In particular, since $A_k \subset G$, we have $m(A_k) < \alpha$. But $\alpha > 0$ was arbitrary, so we must have $m(A_k) = 0$.
Moreover, $k$ was arbitrary, so $m(A \cap [-k,k]) = 0$ for every $k$. Since $A = \bigcup_{k = 1}^\infty (A \cap [-k,k])$, by countable additivity we conclude $m(A) = 0$.
Now let's drop the assumption $m(A) < \infty$. For any $n$ and any interval $I$, we have $$m(A \cap [-n,n] \cap I) \le m(A \cap I) \le (1-\epsilon)m(I).$$
Hence $A \cap [-n,n]$ satisfies the same condition and moreover $m(A \cap [-n,n]) \le 2n < \infty$. So by the previous case, $m(A \cap [-n,n]) = 0$. Now since $A = \bigcup_{n=1}^\infty (A \cap [-n,n])$, by countable additivity $m(A) = 0$.
For an explicit example with a fat Cantor set, let's consider the example given on Wikipedia, where at stage $n \ge 1$ we remove $2^{n-1}$ intervals, each of length $2^{-2n}$. The final set $C$ has measure $1 - \sum_{n=1}^\infty 2^{n-1} \cdot 2^{-2n} = 1 - \sum_{n=1}^\infty 2^{-n-1} = \frac{1}{2}$.
Let $I_k$ be the leftmost interval that remains after stage $k$. At stage $k$ the leftmost interval that was removed was centered at $2^{-k}$ and had length $2^{-2k}$, so the leftmost interval that remains is
$$I_k = \left[0, 2^{-k} - \frac{1}{2} 2^{-2k}\right] = [0, 2^{-k}(1-2^{-k-1})].$$
At the next stage, we will remove from $I_k$ one interval of length $2^{-2(k+1)}$ from $I$, then two intervals of length $2^{-2(k+2)}$ and so on. So the total length of the intervals removed from $I_k$ is
$$\sum_{n=1}^\infty 2^{n-1} 2^{-2(k+n)} = 2^{-2k} \sum_{n=1}^\infty 2^{-n-1} = 2^{-2k} \frac{1}{2} = 2^{-2k-1}.$$
Therefore, we have
$$\frac{m(C \cap I_k)}{m(I_k)} = \frac{m(I_k) - 2^{-2k-1}}{m(I_k)} = \frac{2^{-k}(1-2^{-k-1}) - 2^{-2k-1}}{2^{-k}(1-2^{-k-1})} = \frac{(1-2^{-k-1}) - 2^{-k-1}}{(1-2^{-k-1})}$$
upon cancelling a factor of $2^{-k}$. It is clear by inspection that $$\lim_{k \to \infty} \frac{m(C \cap I_k)}{m(I_k)} = 1,$$ so given $\epsilon > 0$ you can choose $k$ so large that $m(C \cap I_k) > (1-\epsilon) m(I_k)$.
Let $0.a_1a_2a_3a_4\ldots$ be a number in $[0,1]$. If the $n^{th}$ digit is the first 9 in the expansion, there are $9^{n-1}$ possible assignments of digits ($0,1,\ldots,8$) to $a_1, a_2, \ldots, a_{n-1}$. For each of these assignments, we have an interval of length $\frac{1}{10^n}$ of numbers having a nine in their expansions.
Summing up over $n$, we get our answer: $$\lambda(A)=\sum_{n=1}^{\infty}\frac{9^{n-1}}{10^n}=\frac{1}{9}\sum_{n=0}^{\infty}\Big(\frac{9}{10}\Big)^n-\frac{1}{9}=\frac{1}{9}\cdot\frac{1}{1-\frac{9}{10}}-\frac{1}{9}=1$$
Best Answer
See, study the complement of the set. That is, look at integers which contain $5$ in their decimal expansion. Caveat : note that $0.6 = 0.5\overline{9}$ also counts as a decimal which is expressed with a $5$ as one of the digits, so belongs in the set. In particular, any terminating decimal which terminates with $6$ can be considered to belong to the set.
The first such category we can think of is: Those that have $5$ as the first digit following the decimal point. This is the set of numbers $[0.5,0.6]$. This has measure $0.1$, so the left over measure is $0.9$.
Now, from the remaining set, remove the set of all numbers with second digit $5$. This consists of $[0.05,0.06[, [0.15,0.16] \ldots [0.95,0.96]$ without $[0.55,0.56]$, since that was already removed earlier. Now, each of these has measure $0.01$, so we have removed $0.09$ more from the system. Hence, the left over is $0.81$.
By induction, prove the following : at the $n$th step, the set left over has measure $\frac{9^n}{10^n}$. Now, as $n \to \infty$, we see that the given set has measure zero (I leave you to rigorously show this, you can use the Borel-Cantelli lemma). This also incorporates the fact that the given set is measurable, since it's measure is computable(and is $0$).