Term-by-term_Continuity_Theorem

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Definition

Let $(f_n)$ be a sequence of continuous functions on a set $A \subseteq \mathbb{R}$. If $(f_n) \to f$ uniformly on $A$, then $f$ is continuous on $A$.

Proof

Let $c \in A$ and $\epsilon > 0$.

By uniform convergence, there exists $N$ such that $|f_n(x) - f(x)| < \epsilon/3$ for all $x \in A$ and all $n \geq N$.

Since $f_N$ is continuous at $c$, there exists $\delta > 0$ such that $|x - c| < \delta$ implies $|f_N(x) - f_N(c)| < \epsilon/3$.

For $|x - c| < \delta$:

$$ |f(x) - f(c)| \leq |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)| $$$$ < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $$

Thus $f$ is continuous at $c$.