I'll try to explain it in linear case.
Consider the linear model $$Y_i=\sum_{j=1}^{p} \beta_jX_{i}^{(j)}+\epsilon_i, i=1,...,n. $$
When $p \leq n$ (number of independent variables less or equal then number of observation) and design matrix has full rank, the least squared estimator of $b$ is $$\hat{b}=(X^TX)^{-1}X^TY$$ and prediction error is $$ \dfrac{\| X(\hat{b}-\beta^0) \|_2^2}{\sigma^2} $$
from which we can deduce $$ \dfrac{ \mathbb{E} \| X(\hat{b}-\beta^0) \|_2^2}{n}=\dfrac{\sigma^2}{n}p. $$
It means that each parameter $\beta_j^0$ is estimated with squared accuracy $\sigma^2/n, j=1,...,p.$ So your overall squared accuracy is $(\sigma^2/n)p.$
Now what if the number of observations are less then number of independent variables $(p>n)$? We "believe" that not all of our independent variables play a role in explaining $Y$, so only a few, say $k$, of them are non-zero. If we would know which variables are non-zero, we could neglect all other variables and by above argument, overall squared accuracy would be $(\sigma^2/n)k.$
Because the set of nonzero variables are unknown, we need some regularization penalty(for example $l_1$) with regularization parameter $\lambda$ (which controls the number of variables). Now you want to get results similar to discussed above, you want to estimate squared accuracy. The problem is your optimal estimator $\hat{\beta}$ is now depend on $\lambda$. But the great fact is that with proper choice for $\lambda$ you can get an upper bound of prediction error with high probability, that is the "oracle inequality" $$\dfrac{\| X(\hat{\beta}-\beta^0) \|_2^2}{n} \leq const.\dfrac{\sigma^2\log p}{n}k. $$
Note an additional factor $\log p$, which is the price for not knowing set of non-zero variables. "$const.$" depend only on $p$ or $n$.
