# Counterexamples

## Non-integrable functions

• $$\int {1\over 1 + x^2} = \arctan(x) \overset{x\to\infty}\to \pi/2 < \infty$$

• Any bounded function (or continuous on a compact set, by EVT)

• $$\int_0^1 {1 \over \sqrt{x}} < \infty$$

• $$\int_0^1 {1\over x^{1-{\varepsilon}}} < \infty$$

• $$\int_1^\infty {1\over x^{1+{\varepsilon}}} < \infty$$

• $$\int_0^1 {1\over x} = \infty$$.
• $$\int_1^\infty {1\over x} = \infty$$.
• $$\int_1^\infty {1 \over \sqrt{x}} = \infty$$
• $$\int_1^\infty {1\over x^{1-{\varepsilon}}} = \infty$$
• $$\int_0^1 {1\over x^{1+{\varepsilon}}} = \infty$$

Sequences $$f_k \overset{a.e.}\to f$$ but $$f_k \overset{L^p}{\not\to} f$$:

• For $$1\leq p < \infty$$: The skateboard to infinity, $$f_k = \chi_{[k, k+1]}$$.

Then $$f_k \overset{a.e.}\to 0$$ but $${\left\lVert {f_k} \right\rVert}_p = 1$$ for all $$k$$.

Converges pointwise and a.e., but not uniformly and not in norm.

• For $$p = \infty$$: The sliding boxes $$f_k = k \cdot \chi_{[0, \frac 1 k]}$$.

Then similarly $$f_k \overset{a.e.}\to 0$$, but $${\left\lVert {f_k} \right\rVert}_p = 1$$ and $${\left\lVert {f_k} \right\rVert}_\infty = k \to \infty$$

Converges a.e., but not uniformly, not pointwise, and not in norm.

Notions of convergence:

• Uniform
• Pointwise
• Almost everywhere
• In norm

Uniform $$\implies$$ pointwise $$\implies$$ almost everywhere, but in general non of these can be reversed.

• Uniform: $$f_n \rightrightarrows f: \forall \varepsilon ~\exists N {~\mathrel{\Big\vert}~}~n\geq N \implies {\left\lvert {f_N(x) - f(x)} \right\rvert} < \varepsilon \quad \forall x.$$
• Pointwise: $$f_n(x) \to f(x)$$ for all $$x$$. (This is just a sequence of numbers)
• Almost Everywhere: $$f_n(x) \to f(x)$$ for almost all $$x$$.
• Norm: $${\left\lVert {f_n - f} \right\rVert}_1 = \int {\left\lvert {f_n(x) - f(x)} \right\rvert} \to 0$$.

We have $$1 \implies 2 \implies 3$$, and in general no implication can be reversed, but (warning) none of $$1,2,3$$ imply $$4$$ or vice versa.

• $$f_n = (1/n) \chi_{(0, n)}$$. This converges uniformly to 0, but the integral is identically 1. So this satisfies 1,2,3 and not 4.

• $$f_n = \chi_{(n, n+1)}$$ (skateboard to infinity). This satisfies 2,3 but not 1, 4.

• $$f_n = n\chi_{(0, \frac 1 n)}$$. This satisfies 3 but not 1,2,4.

• $$f_n:$$ one can construct a sequence where $$f_n \to 0$$ in $$L^1$$ but is not 1,2, or 3. The construction:

• Break $$I$$ into $$2$$ intervals, let $$f_1$$ be the indicator on the first half, $$f_2$$ the indicator on the second.
• Break $$I$$ into $$2^2=4$$ intervals, like $$f_3$$ be the indicator on the first quarter, $$f_4$$ on the second, etc.
• Break $$I$$ into $$2^k$$ intervals and cyclic through $$k$$ indicator functions.

• Then $$\int f_n = 1/2^n \to 0$$, but $$f_n\not\to 0$$ pointwise since for every $$x$$, there are infinitely many $$n$$ for which $$f_n(x) = 0$$ and infinitely many for which $$f_n(x) = 1$$.

Almost everywhere convergence does not imply $$L^p$$ convergence for any $$1\leq p \leq \infty$$. In the following examples, $$f_k \overset{a.e.}\to f$$ but $$f_k \overset{L^p}{\not\to} f$$:

• For $$1\leq p < \infty$$: The skateboard to infinity, $$f_k = \chi_{[k, k+1]}$$.

• Then $$f_k \overset{a.e.}\to 0$$ but $${\left\lVert {f_k} \right\rVert}_p = 1$$ for all $$k$$.
• Converges pointwise and a.e., but not uniformly and not in norm.
• For $$p = \infty$$: The sliding boxes $$f_k = k \cdot \chi_{[0, \frac 1 k]}$$.

• Then similarly $$f_k \overset{a.e.}\to 0$$, but $${\left\lVert {f_k} \right\rVert}_p = 1$$ and $${\left\lVert {f_k} \right\rVert}_\infty = k \to \infty$$

• Converges a.e., but not uniformly, not pointwise, and not in norm.

For any measure space $$(X, {\mathcal{M}}, \mu)$$,

• $$L^1(X)$$ is Banach space.
• $$L^2(X)$$ is a (possibly non-separable) Hilbert space.