- #1
Castilla
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Hello guys, I am following the chapter about multiple Riemann integrals in Apostol's Mathematical Analysis. Theorem 14.11 says this (I translate from spanish to english):
"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
"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