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[[File:Cantor function sequence.png|250px|right]]
Below we define a sequence {''ƒf''<sub>''n''</sub>} of functions on the unit interval that converges to the Cantor function.
Let ''ƒf''<sub>0</sub>(''x'') = ''x''.
Then, for every integer {{nowrap|''n'' ≥ 0}}, the next function ''ƒf''<sub>''n''+1</sub>(''x'') will be defined in terms of ''ƒf''<sub>''n''</sub>(''x'') as follows:
Let ''ƒf''<sub>''n''+1</sub>(''x'') = {{nowrap|1/2 × ''ƒf''<sub>''n''</sub>(3''x'')}}, when {{nowrap|0 ≤ ''x'' ≤ 1/3 }};
Let ''ƒf''<sub>''n''+1</sub>(''x'') = 1/2, when {{nowrap|1/3 ≤ ''x'' ≤ 2/3 }};
Let ''ƒf''<sub>''n''+1</sub>(''x'') = {{nowrap|1/2 + 1/2 × ''ƒf''<sub>''n''</sub>(3 ''x'' − 2)}}, when {{nowrap|2/3 ≤ ''x'' ≤ 1}}.
The three definitions are compatible at the end-points 1/3 and 2/3, because ''ƒf''<sub>''n''</sub>(0) = 0 and ''ƒf''<sub>''n''</sub>(1) = 1 for every ''n'', by induction. One may check that ''ƒf''<sub>''n''</sub> converges pointwise to the Cantor function defined above. Furthermore, the convergence is uniform. Indeed, separating into three cases, according to the definition of ''ƒf''<sub>''n''+1</sub>, one sees that
:<math>\max_{x \in [0, 1]} |f_{n+1}(x) - f_n(x)| \le \frac 1 2 \, \max_{x \in [0, 1]} |f_{n}(x) - f_{n-1}(x)|, \quad n \ge 1.</math>
If ''ƒf'' denotes the limit function, it follows that, for every ''n'' ≥ 0,
:<math>\max_{x \in [0, 1]} |f(x) - f_n(x)| \le 2^{-n+1} \, \max_{x \in [0, 1]} |f_1(x) - f_0(x)|.</math>
Also the choice of starting function does not really matter, provided ''ƒf''<sub>0</sub>(0) = 0, ''ƒf''<sub>0</sub>(1) = 1 and ''ƒf''<sub>0</sub> is [[Bounded function|bounded]]{{citation needed|date=September 2014}}.
=== Fractal volume ===
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