Let $K\subset \mathbb C$ be a closed subset of the complex plane, not necessarily bounded.  
 Let $U$ be the interior of $K$.


Let $f:K\to \mathbb C$ be a continuous bounded function, whose restriction to $U$ is holomorphic.


Assume furthermore that for every closed curve $\gamma\subset K$, the integral $\int\_\gamma f(z)dz$ vanishes. (this is only relevant if $K$ is not simply connected). If you prefer, you can assume that $K$ is simply connected.



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> Can $f$ be uniformly approximated by entire functions?
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Is Every Holomorphic Near an Entire?
As mentioned in the comments, this is true if $K$ is compact and the complement of $K$ in the Riemann sphere is connected : it is the content of Mergelyan's Theorem on uniform polynomial approximation of holomorphic functions.


**EDIT** If $K$ is only assumed to be closed, this is also true with the additional assumption that $K$ is locally connected at infinity : it is the content of Arakelian's theorem on the uniform approximation of holomorphic functions by entire functions.


It is false in general if the complement of $K$ is not connected : consider $K$ the annulus $\{1/2 \leq |z| \leq 2\}$ and $f(z):=1/z^2$. Then $f$ is continuous on $K$, holomorphic in the interior of $K$, and $\int\_\gamma f(z)\, dz=0$ for all closed curve $\gamma$ contained in $K$. However, $f$ cannot be uniformly approximated on $K$ by entire functions $(f\_n)$. Indeed, if it were the case, then the functions $g\_n(z):=zf\_n(z)$ would uniformly approximate $g(z):=zf(z)=1/z$ on $K$. This contradicts the fact that $\int\_\mathbb{T} g\_n(z)\, dz=0$ for all $n$, whereas $\int\_\mathbb{T} g(z) \, dz = 2\pi i \neq 0$. Here $\mathbb{T}$ is the unit circle. 


