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1d
revised Approximating functions such that the left- and right limit exists everywhere
added 31 characters in body
1d
asked Approximating functions such that the left- and right limit exists everywhere
2d
accepted If a function is enclosed by lower and upper sums, does its limit w.r.t. partitions equals the integral
2d
comment Bounded function on compact interval that is not Lebesgue integrable
How then is the sentence "In the general no example existed for a bounded functions with compact support to be not lebesgue integrable" above to be meant?
2d
comment If a function is enclosed by lower and upper sums, does its limit w.r.t. partitions equals the integral
Yes, found a better notation!
2d
revised If a function is enclosed by lower and upper sums, does its limit w.r.t. partitions equals the integral
added 10 characters in body
2d
asked If a function is enclosed by lower and upper sums, does its limit w.r.t. partitions equals the integral
Jun
27
comment Reasoning about numbers close to two other numbers $a,b$ (inequalities)
Just for the record, another solution, starting by trying to show that $|x-y|$ subtracted gives less than $\varepsilon$ substracted, thereby $|x-y|$ must be bigger: $$ (b - a) - | x - y | \le | (b - a) - (y - x) | = | (x - a) + (b - y) | \le | x - a | + |y - b| \le (b-a) - \varepsilon $$ and hence $ -|x - y| \le -\varepsilon \Leftrightarrow |x-y| \ge \varepsilon$.
Jun
27
accepted Reasoning about numbers close to two other numbers $a,b$ (inequalities)
Jun
27
comment Reasoning about numbers close to two other numbers $a,b$ (inequalities)
Yes, thank you! I meant it differently. I edited my question, the remaining length $(b-a) - \varepsilon$ has to be divided by two.
Jun
27
revised Reasoning about numbers close to two other numbers $a,b$ (inequalities)
added 20 characters in body
Jun
27
asked Reasoning about numbers close to two other numbers $a,b$ (inequalities)
Jun
26
comment Bounded function on compact interval that is not Lebesgue integrable
Thank for this very interesting additional information. Do you have a proof that a bounded function with compact support is Lebesgue-measurable?
Jun
26
accepted Bounded function on compact interval that is not Lebesgue integrable
Jun
26
comment Approximation Lemma for Riemann-integrable functions
Yes, a function is Riemann-integrable iff its set of discontinuities has measure zero, but I am looking more for some approximation by sequences.
Jun
26
asked Approximation Lemma for Riemann-integrable functions
Jun
26
asked Bounded function on compact interval that is not Lebesgue integrable
Jun
26
accepted Evaluate $ \int_{\varepsilon}^1 \sin\left( \frac{1}{x} \right) dx$
Jun
26
comment Evaluate $ \int_{\varepsilon}^1 \sin\left( \frac{1}{x} \right) dx$
@Airbag: I could not find it...
Jun
26
asked Evaluate $ \int_{\varepsilon}^1 \sin\left( \frac{1}{x} \right) dx$