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Suppose that we take some continuous real function of a real variable $f$ which is defined on the set $[a,b]$ and that we have $\int_{a}^{b}f(x)dx=\alpha$.

Now, let us denote by $SIV_{\alpha}([a,b])$ the set of functions such that we have, if $g \in SIV_{\alpha}([a,b])$ then $g$ is defined on $[a,b]$, $g$ is continuous on $[a,b]$ and $\int_{a}^{b}g(x)dx=\alpha$.

So $SIV_{\alpha}([a,b])$ is the set of all real functions of a real variable which are defined and continuous on the set $[a,b]$ and whose integral on the set $[a,b]$ is equal $\alpha$.

Let now $C([a,b])$ be the set of all continuous real functions of a real variable defined on the set $[a,b]$.

Does there exist bijection between the sets $C([a,b])$ and $SIV_{\alpha}([a,b])$?

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The cardinality of $SIV_{\alpha}([a,b])$ is equal to the one of $C([a,b])$ which is equal to the one of $\mathbb R$, i.e. the cardinality of the continuum.

To see that $C([a,b])$ has the cardinality of the continuum, you can have a look to this post.

Then it is easy to see that $SIV_{\alpha}([a,b])$ has also at least the cardinality of the continuum as you can find $f \in SIV_{\alpha}([a,b])$ with $f(a)$ equal to whatever real number you want. As the cardinality of $SIV_{\alpha}([a,b])$ is obviously less than or equal to the one of $C([a,b])$, you're done.

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  • I wonder if it's possible to exhibit an explicit bijection between the two sets. I suppose it must be, because I can't see the Axiom of Choice anywhere here. (PS I think "obviously less than" should be "obviously less than or equal to".) – TonyK Dec 08 '15 at 17:06
  • Using the Cantor Bernstein theorem which has a constructive version, we just have to find a one-to-one map from $SIV_\alpha ([a, b]) $ into $C([a, b]) $. – mathcounterexamples.net Dec 08 '15 at 17:23
  • Unfortunately, in English you really do have to say "less than or equal to". "Less or equal than" doesn't cut it. – TonyK Dec 08 '15 at 17:28
  • Thanks for the English lesson! Not so easy to be precise in a different language as there are a lot of tricky differences... That change mathematical meaning. – mathcounterexamples.net Dec 08 '15 at 17:32