Does the function need to be Lipschitz or is Lebesgue integrable enough?

Let $f:(0,1) to mathbb{R}$, $mathbb{Z}_n ={0,1,dots,n-1}$ and $delta >0$.

Which is the least I can assume about $f$ so that

begin{equation*}
begin{split}
left| frac{1}{n} sum_{ x in mathbb{Z}_n} f(tfrac{x}{n}) – int_{(0,1)}f(u) du right| leq delta
end{split}
end{equation*}

for sufficiently large $n$?

I know for a fact that if $f$ is Lipschitz continuous with constant $K$, then the inequality holds as

begin{equation*}
begin{split}
left| frac{1}{n} sum_{ x in mathbb{Z}_n} f(tfrac{x}{n}) – int_{(0,1)}f(u) du right| & = left| frac{1}{n} sum_{ x in mathbb{Z}_n} f(tfrac{x}{n}) – sum_{x in mathbb{Z}_n}int_{tfrac{x}{n}}^{tfrac{x+1}{n}}f(u) du right| \
& = left| frac{1}{n} sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} (f(tfrac{x}{n}) – f(u) ) du right|\
& leq sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} left|f(tfrac{x}{n}) – f(u)right| du \
& leq sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} K left| tfrac{x}{n}- u right| du leq K sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} tfrac{1}{n} du\
& = K sum_{ x in mathbb{Z}_n} tfrac{1}{n^2}= frac{K}{n}.
end{split}
end{equation*}

So I guess my question is: Can I assume some (relevant) weaker condition about $f$ so that the inequality holds? In particular, is the inequality true if $f$ is just integrable?

analysis – How do I check if the function is integrable (Lebesgue) on R?

Is f(x) = $ sum_{n=0}^{infty}((-1)^{n}/n) X_{(n,n+1)} $ Lebesgue integrable on R? X is the characteristic function.

I’m confused about how to handle the sum and the alternating series part. The sum is clearly conditionally convergent. But it might not be non-negative. Is it still integrable over R? (Lebesgue).

New here, sorry for the crap formatting.

partial differential equations – If $S(t)$ is a $C_0$-semigroup, is $S(t-s)f(s)$ Bochner integrable?

Let $X$ be a Banach space and let $S(t)$, $t geq 0$, be a $C_0$-semigroup on $X$.

Assume that $f : (0,+infty) rightarrow X$ is Bochner integrable.

Is $S(t-s)f(s)$ Bochner integrable on $(0,t)$ and does $t mapsto int_0^t S(t-s)f(s)ds in C^0((0,+infty),X)$ ?

The function $t mapsto int_0^t S(t-s)f(s)ds$ arises when we define the notion of weak solution to an inhomogeneous evolution PDE $$partial_t u(t) = Au(t) + f(t), quad u(0) = u_0$$

where $A$ is the infinitesimal generator of $S(t)$.

If $f$ is continuous, I know that the result is true, but I’m interested in the non-continuous case. I would expect this to be true.

Any proof or reference is welcomed.

real analysis – Interval of $p$ that makes the $p$ power of a function integrable

For a measurable real-valued function $f$ on $(0,infty)$ let $P(f):={pin(0,infty):|f|^pinmathcal{L}^1(mathbb{R},mathcal{B},lambda)}$, where $mathcal{B}$ is the Borel $sigma$-algebra. Show that for every subinterval $J$ of $(0,infty)$, which may be open or closed on the left and on the right, there is some $f$ such that $P(f)=J$. Hint: Consider functions such as $x^a|log x|^b$ on $(0,1)$ and on $(1,infty)$, where $b=0$ unless $a=-1$.

I don’t understand the hint. $f(x)=x^anotinmathcal{L}^1(mathbb{R},mathcal{B},lambda)$ for any $a$. The log part is also not integrable:

begin{equation}
begin{split}
& int_0^infty x^{-1}|log x|^bdx \
= & int_0^infty |log x|^bdlog x \
= & int_0^1 (-log x)^bdlog x+int_1^infty (log x)^bdlog x \
= & int_{-infty}^0 (-x)^bdx+int_0^infty x^bdx \
= & -int_{-infty}^0 (-x)^bd(-x)+int_0^infty x^bdx \
= & -int_infty^0 x^bdx+int_0^infty x^bdx \
= & int_0^infty x^bdx+int_0^infty x^bdx \
end{split}
end{equation}

But for $J=(0,infty)$, I can find such a $f(x)$:
begin{equation}
f(x) = begin{cases}
x-1 &text{$1<xleq2$}\
3-x &text{$2<xleq3$}\
0 &text{otherwise}
end{cases}
end{equation}

real analysis – A necessary and sufficient condition of Lebesgue integrable.

Let $lgeqq 2$ be natural number.
And let $f$ be a Lebesgue-measurable function and suppose $f$ is non negative.

Prove that $f$ is Lebesgue-integrable if and only if $displaystylesum_{n=-infty}^{infty} l^n cdot m(A_n)$ converges where $m$ is Lebesgue measure and $A_n={xin mathbb{R^n} | l^n leqq f(x)<l^{n+1} }.$

I’m not sure my proof is correct.

My proof

First, suppose $displaystylesum_{n=-infty}^{infty} l^n cdot m(A_n)$ converges.

Let $A=displaystylebigcup_{n=-infty}^{infty} A_n.$

Note that $displaystyleint_{bigcup_{n=-infty}^{infty} A_n} f(x) dx=sum_{n=-infty}^{infty} int_{A_n} f(x) dx$ since $A_jcap A_k=emptyset$ for $jneq k$ and $A^c=emptyset$ since $A={ xin mathbb{R^n} | f(x)geqq 0}.$

Then,
begin{align}
displaystyleint |f(x)| dx&=displaystyleint f(x) dx\
&=int_{A} f(x) dx +int_{A^c} f(x) dx\
&=int_{bigcup_{n=-infty}^{infty} A_n} f(x) dx +int_{emptyset} f(x) dx\
&=sum_{n=-infty}^{infty} int_{A_n} f(x) dx +0\
&<sum_{n=-infty}^{infty} int_{A_n} l^{n+1} dx \
&=sum_{n=-infty}^{infty} l^{n+1} cdot m(A_n) \
&=l sum_{n=-infty}^{infty} l^{n} cdot m(A_n) \
&<infty.
end{align}

Thus $f$ is Lebesgue-integrable.

Conversely, suppose $f$ is Lebesgue-integrable.

Then,
begin{align}
sum_{n=-infty}^{infty} l^{n} cdot m(A_n)
&=sum_{n=-infty}^{infty} l^{n} displaystyleint_{A_n} dx \
&=sum_{n=-infty}^{infty} displaystyleint_{A_n} l^{n} dx \
&leqq sum_{n=-infty}^{infty} displaystyleint_{A_n} f(x) dx \
&=displaystyleint_{bigcup_{n=-infty}^{infty} A_n} f(x) dx \
&=displaystyleint_{A} f(x) dx \
&=displaystyleint_{A} f(x) dx +displaystyleint_{emptyset} f(x) dx \
&=displaystyleint_{A} f(x) dx +displaystyleint_{A^c} f(x) dx \
&=displaystyleint f(x) dx \
&=displaystyleint |f(x)| dx \
&<infty.
end{align}

Is this proof correct?

real analysis – Prove that if $ f $ is Riemann Integrable then it is bounded

I get the idea of how to do this proof, essentially I would choose a pathological partition $P$ such that it includes a point $ c in P $ where $ f(c) Delta_c – l geq epsilon$. However I’m stuck on how to make this work in a $ delta,epsilon $ proof.

Suppose the contrary that $ f $ is Riemann Integrable but $ f $ is not bounded. It suffices I show

$$ exists epsilon > 0, forall delta > 0 text{s.t} exists P, m(P) < delta mathrm{but} |R(f,P) – l| geq epsilon $$

  • $ m(P) $ is the mesh of the partition $P$

Since $ f $ is not bounded, choose $ c $ such that $ f(c) > frac{l + epsilon}{delta} $ and let $ c in P $

Then I initially I had this:

$$ |R(f,P) – l| = left|sum_{{s_j} in P} f(s_j) Delta_j – lright| $$
$$ geq left|sum_{{s_j} in P} f(s_j) delta – lright|$$
$$ geq left|f(c) delta – lright|$$
$$ geq left|frac{l + epsilon}{delta} delta – lright|$$
$$ = epsilon $$

However, clearly the inequality here is nowhere near correct, since the difference is an absolute value. How would I go about this proof?

(I’m following the Stephen Krantz, Real Analysis and Foundations Textbook)

measure theory – Bochner integrability of composition of continuous and Bochner integrable function

Let $(varOmega, varSigma ,mu)$ be a finite measure space and $X$ be a separable Banach space. For $1 leq p leq infty$, let $L_{p}(mu,X)$ denote the class of all $mu$-Bochner integrable functions $f$ such that $|f|_{p} = (int_{varOmega} |f|^{p})^{1/p} < infty$.

If $Phi in L_{p}(mu,X)$ and $h: X rightarrow X$ is a continuous function, then is $h circ Phi in L_{p}(mu,X)$?

calculus – Determine if the function $f(x)$ is integrable in $[0,1]$

Determine if the function $f(x)$ is integrable in $(0,1)$

$$displaystyle f( x) =begin{cases}
frac{1}{x^{2}} & x >0\
0 & x=0
end{cases}$$


Attempt:

Let’s check the limits of each side $0^+,0^-$ to show that $frac{1}{x^2}$ is not bounded.

$$displaystyle lim _{xrightarrow 0^{+}}frac{1}{x^{2}} =infty $$

$$displaystyle lim _{xrightarrow 0^{-}}frac{1}{x^{2}} =infty $$

Therefore, the function is not bounded when $lim_{xrightarrow 0^{+/-}}$
, and is not integrable in $(0,1)$

In addition, there is a well-known theorem:

If $f$ is a continuous function except for the finite number of unconsciously points $Rightarrow$ $f$ is integrable.

Which contradicts the attempt.

real analysis – Proving f Riemann integrable implies |f| Riemann integrable by contradiction.

I know that there exists a theorem telling us that if a real valued function $f$ defined over an interval $(a,b)$ is Riemann integrable, then |$f$| is too (moreover I believe the converse to be false). I’ve tried to look at the proofs, but they seem a bit beyond my ability at the moment. My first thought was to suppose there existed a Riemann integrable function, let us say $g$, such that |$g$| was not Riemann integrable, and show that this leads to some sort of contradiction.

I thought about this for a while and have not been able to think of such a function/proof. Would anybody know of such an example?

real analysis – prove $log x$ is integrable near the origin

Prove that $log(x) in L^1((0,1))$.which has been shown in this post.

The first idea is using the integral $int_epsilon^1 log(x) dx$ to approximate it.which may be the definition in classical analysis.I was bit confused for why this limit: $lim_{epsilon to 0} I(epsilon) = int_0^1 log x dx$ holds in Lebesgue integral. Maybe we need to take an absolute value first $int_epsilon^1 |log(x)|dx$ first then using the Monotone convergence theorem for non-negative function, rest of the proof are the similar ,is my interpretation correct?

The second idea also shows in this post:using the fact that $x^{-alpha} in L^1((0,1))$ for $alpha <1$ then near the origin ,we have $x^{alpha}log x to 0$ for all $alpha >0$ which means exist a small neiborhood near origin says $(0,delta)$ such that if $xin (0,delta)$ we have $|x^alog x|le 1/2$ i.e. $|log x|le frac{1}{2}x^{-a}$ since the RHS is integrable hence log is also integrable near origin,is my interpretation correct?