1108: For Interval on R and Continuous Function on It Whose Integral Is Finite, a Way of Changing Function to Be Continuous with Desired Integral

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description/proof of that for interval on $\mathbb{R}$ and continuous function on it whose integral is finite, a way of changing function to be continuous with desired integral

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Topics

About: measure space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of Lebesgue integral on measure space.
• The reader knows a definition of continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that for any interval on $\mathbb{R}$ and any continuous function on it whose integral is finite, a way of changing the function to be continuous with any desired integral is this.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$\mathbb{R}$: $=\text{ the Euclidean topological space }$
$(I,\sigma ,\lambda )$: $\in \{\text{ the intervals of }\mathbb{R}\}$, with the subspace topology, with the Borel $\sigma$-algebra and the Lebesgue measure
$f$: $:I\to \mathbb{C}$, $\in \{\text{ the continuous functions }\}$ such that ${\int }_{I}fd\mu =c<\mathrm{\infty }$
$[x_0,x_1]$: $\subseteq I$
$c^{\prime }$: $\in \mathbb{C}$
$\delta$: $=4(c^{\prime }-c)/(x_1-x_0{)}^2$
$f^{\prime }$: $:I\to \mathbb{C}$ such that on $I\setminus [x_0,x_1]$, $f^{\prime }=f$, on $[x_0,(x_0+x_1)/2]$, $f^{\prime }=f+\delta (x-x_0)$, and on $((x_0+x_1)/2,x_1]$, $f^{\prime }=f+\delta (x_1-x)$
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Statements:
$f^{\prime }\in \{\text{ the continuous functions }\}$
$\wedge$
${\int }_I{f}^{\prime }d\lambda =c^{\prime }$
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2: Note

Of course, this is not the only way.

And of course, this is a quite natural way that anyone will probably hit upon as one of some options, but as we do not want to the same calculation twice, we will record the result here.

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3: Proof

Whole Strategy: Step 1: see that $f^{\prime }$ is continuous; Step 2: see that ${\int }_I{f}^{\prime }d\lambda =c^{\prime }$.

Step 1:

Let us see that $f^{\prime }$ is indeed continuous.

On $I\setminus [x_0,x_1]$, $f^{\prime }$ is continuous, because $f^{\prime }=f$ there.

On $[x_0,(x_0+x_1)/2]$, $f^{\prime }$ is obviously continuous.

On $((x_0+x_1)/2,x_1]$, $f^{\prime }$ is obviously continuous.

The concerns are whether that 1) when $x$ approaches $x_0$ from left, $f^{\prime }(x)$ approaches $f^{\prime }(x_0)$; 2) when $x$ approaches $(x_0+x_1)/2$ from right, $f^{\prime }(x)$ approaches $f^{\prime }((x_0+x_1)/2)$; 3) when $x$ approaches $x_1$ from right, $f^{\prime }(x)$ approaches $f^{\prime }(x_1)$.

For 1), $f^{\prime }(x_0)=f(x_0)+\delta (x_0-x_0)=f(x_0)$, and certainly, when $x$ approaches $x_0$ from left, $f^{\prime }(x)$ approaches $f(x_0)=f^{\prime }(x_0)$.

For 2), $f^{\prime }((x_0+x_1)/2)=f((x_0+x_1)/2)+\delta ((x_0+x_1)/2-x_0)=f((x_0+x_1)/2)+\delta ((x_1-x_0)/2)$, and when $x$ approaches $(x_0+x_1)/2$ from right, $f^{\prime }(x)$ approaches $f((x_0+x_1)/2)+\delta (x_1-(x_0+x_1)/2)=f((x_0+x_1)/2)+\delta ((x_1-x_0)/2)=f^{\prime }((x_0+x_1)/2)$.

For 3), $f^{\prime }(x_1)=f(x_1)+\delta (x_1-x_1)=f(x_1)$, and when $x$ approaches $x_1$ from right, $f^{\prime }(x)$ approaches $f(x_1)=f^{\prime }(x_1)$.

So, $f^{\prime }$ is continuous.

Step 2:

Let us see that indeed ${\int }_I{f}^{\prime }d\lambda =c^{\prime }$.

${\int }_I{f}^{\prime }d\lambda ={\int }_{I\setminus [x_0,x_1]}{f}^{\prime }d\lambda +{\int }_{[x_0,(x_0+x_1)/2]}{f}^{\prime }d\lambda +{\int }_{((x_0+x_1)/2,x_1]}{f}^{\prime }d\lambda ={\int }_{I\setminus [x_0,x_1]}fd\lambda +{\int }_{[x_0,(x_0+x_1)/2]}(f+\delta (x-x_0))d\lambda +{\int }_{((x_0+x_1)/2,x_1]}(f+\delta (x_1-x))d\lambda ={\int }_{I\setminus [x_0,x_1]}fd\lambda +{\int }_{[x_0,(x_0+x_1)/2]}fd\lambda +{\int }_{((x_0+x_1)/2,x_1]}fd\lambda +{\int }_{[x_0,(x_0+x_1)/2]}\delta (x-x_0)d\lambda +{\int }_{((x_0+x_1)/2,x_1]}\delta (x_1-x)d\lambda ={\int }_{I}fd\lambda +{\int }_{[x_0,(x_0+x_1)/2]}\delta (x-x_0)d\lambda +{\int }_{((x_0+x_1)/2,x_1]}\delta (x_1-x)d\lambda =c+{\int }_{[x_0,(x_0+x_1)/2]}\delta (x-x_0)d\lambda +{\int }_{((x_0+x_1)/2,x_1]}\delta (x_1-x)d\lambda$.

${\int }_{[x_0,(x_0+x_1)/2]}\delta (x-x_0)d\lambda =\delta [(x-x_0{)}^2/2{]}_{x_0}^{(x_0+x_1)/2}=\delta /2((x_0+x_1)/2-x_0{)}^2=\delta /2((x_1-x_0)/2{)}^2=\delta /8(x_1-x_0{)}^2$.

${\int }_{((x_0+x_1)/2,x_1]}\delta (x_1-x)d\lambda =-\delta [(x_1-x{)}^2/2{]}_{(x_0+x_1)/2}^{x_1}=\delta /2(x_1-((x_0+x_1)/2){)}^2=\delta /2((x_1-x_0)/2{)}^2=\delta /8(x_1-x_0{)}^2$.

So, ${\int }_I{f}^{\prime }d\lambda =c+\delta /8(x_1-x_0{)}^2+\delta /8(x_1-x_0{)}^2=c+\delta /4(x_1-x_0{)}^2=c+4(c^{\prime }-c)/(x_1-x_0{)}^2/4(x_1-x_0{)}^2=c+(c^{\prime }-c)=c^{\prime }$.

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References

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