- #1

Castilla

- 241

- 0

"Let S be a Jordan measurable set. Let the function f be defined and bounded in S. Then f is Riemann integrable if and only if the discontinuities of f in S are a set of cero measure".

The proof says:

Let L be a compact interval that contains S and let g(x) = f(x) when x belongs to S, g(x) = 0 when x belongs to L - S. The discontinuities of the function f will be discontinuities of the function g. But the function g may have also discontinuities en some or all the points of the boundary of S (...).

Now, I understand that there are only two logical options:

- S includes the boundary of S.

- S does not includes the boundary of S.

Now, if S includes the boundary of S then the function f is defined and bounded in the boundary of f and therefore the only discontinuities of g in L are those of f in S.

And if S does not includes the boundary of S, L - S includes it. Then if x is a point in said boundary g(x) is always 0, then g is constant in said boundary, hence g is continuous in said boundary. And so we have the same result that in the other case: the only discontinuities of g in L are those of f in S.

But if this is right, why the proof says that g may also have discontinuities in some or all the points of the boundary of S? Where it is my flaw??

Maybe this is a kind of stupid question. If it is so, I beg your pardon for wasting your time. Nevertheless I dare to ask your help.

Thanks in any case,

Pedro Castilla

Lima, Perú