Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous non-negative function and let $g$ be convex function on $[0,\infty)$. Show that $(g \, o \, f )$ is Riemann Integrable on $[0,1]$.
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$g \circ f$ is continuous. A continuous function on a compact set is uniformly continuous, so it is Riemann integrable. – Tunococ Mar 14 '14 at 04:20
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@math123 Did you get something from my answer below? – Etienne Mar 25 '14 at 17:55
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@math123 Thanks for the bounty! – Etienne Mar 26 '14 at 18:10
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We have that every convex function is continuous. And we know that the composition of continuous functions is continuous. Continuity is sufficient for Riemann integrability.
ml0105
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No, $g$ need not to be continuous. $g$ must be continuous on $(0, \infty)$, but not at 0. For example, the function $g(0)=1$ and $g(x)=x$ if $x>0$ is convex, but not continuous at 0. – math123 Mar 18 '14 at 10:12
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The function $g\circ f$ need not be Riemann-integrable.
To see this, take a closed set $C\subset [0,1]$ which has empty interior but non-zero Lebesgue measure. Let $f:[0,1]\to \mathbb R$ be a continuous function which is identically $0$ on $C$ and (strictly) positive on $[0,1]\setminus C$. for example, one can take $f(x):=dist(x,C)\, .$
Now, define $g:[0,\infty)\to\mathbb R\,$ by $g(0)=1$ and $g(u)=0$ on $(0,\infty)\,$. This function is convex.
By the definitions of $f$ and $g$ we have $g\circ f(x)=1$ if $x\in C$ and $g\circ f(x)=0$ if $x\not\in C$. In other words, $g\circ f$ is the indicator function $\mathbf 1_C$ of $C$. This function is discontinuous at all points $x\in\partial C$, the boundary of $C$, i.e. at all points of $C$ since $C$ is closed and has empty interior. Since $C$ has non-zero Lebesgue measure, it follows that $g\circ f$ is not Riemann-integrable (by Lebesgue's criterion for Riemann-integrability).
Etienne
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