Extendability Theorem for Function Continuous on Open Interval

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Theorem

Let $f$ be a continuous real function that is defined on an open interval $\openint a b$.

Let $g$ be a real function that satisfies:

$g$ is defined on $\closedint a b$
$g$ is continuous on $\closedint a b$
$g = f$ on $\openint a b$.


Then $g$ exists if and only if $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.


Proof

Necessary Condition

Assume that $g$ exists.

We need to prove that the limits $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.


$g$ is continuous at the end points $a$ and $b$ of its domain as $g$ is continuous on $\closedint a b$.

$g$ is right-continuous at $a$ and left-continuous at $b$.

This means that $\map g a = \displaystyle \lim_{x \mathop \to a^+} \map g x$ and $\map g b = \displaystyle \lim_{x \mathop \to b^-} \map g x$.

In turn, this means that the expressions $\displaystyle \lim_{x \mathop \to a^+} \map g x$ and $\displaystyle \lim_{x \mathop \to b^-} \map g x$ exist.

The two $x$ parameters as being part of the two limiting processes in these expressions can be considered confined to $\openint a b$.

Therefore, $g$ in these two expressions can be replaced by $f$ as $g = f$ on $\openint a b$.

We conclude that $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.

$\Box$


Sufficient Condition

Let $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.

We need to prove that a function $g$ with the properties listed in the theorem exists.


Define $g = f$ on $\openint a b$.

Define:

$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map f x$
$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map f x$

$g$ is defined at $a$ and $b$ as ensured by the existence of the limits on the right hand sides of these two equations.

We know that $f$ is continuous on $\openint a b$.

Therefore, $g$ too is continuous on $\openint a b$ as $g = f$ there.


It remains to show that $g$ is continuous at $a$ and $b$.


$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map f x$ per definition.

$x$ can be considered as being an element of $\openint a b$ as the limiting process at $a$ requires $x$ to approach $a$ from above.

$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map f x$ per definition.

$x$ can be considered as being an element of $\openint a b$ as the limiting process at $b$ requires $x$ to approach $b$ from below.

Therefore, the two equations above can be written:

$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map g x$
$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map g x$

as $g = f$ on $\openint a b$.

We conclude that $g$ is continuous at $a$ and $b$ as these two equations are exactly the definitions of continuity for $g$ at respectively $a$ and $b$.

$\blacksquare$