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 Incoming links: Continuous functions are meagre in L1

Past exam of the functional analysis II course of the University of Oxford / 2023 / 2 / b / iv / Solution 2026-10-25

Let

J : C ([0, 1]) \to L^{1} (0, 1)

send a continuous function to its almost everywhere equivalence class. The domain with the supremum norm and the target with the L1 norm are Banach spaces, and

∥ Jf ∥_{L^{1}} = \int_{0}^{1} ∣ f (t) ∣ d t \leq ∥ f ∥_{\infty} .

Thus

J

is a bounded linear operator, and the requested vector subspace is

J (C ([0, 1]))

.

To show this image of a linear map is not closed, consider the continuous functions

f_{n} (t) = min {1, max {0, n (t - 1/2)}} (n \geq 2)

and

g = 1_{(1/2, 1)}

. Their L1 norm difference is the area of one triangle:

∥ J f_{n} - g ∥_{L^{1}} = \int_{1/2}^{1/2 + 1/ n} (1 - n (t - 1/2)) d t = \frac{1}{2 n} ⟶ 0.

But

g

has no continuous function representative. Indeed, a continuous function equal to zero almost everywhere on

(0, 1/2)

must be identically zero there: any nonzero value would persist on an open interval of positive Lebesgue measure. Likewise any continuous representative would be identically one on

(1/2, 1)

. Continuity at

1/2

makes these requirements incompatible. Therefore

g \in / J (C ([0, 1]))

, although it lies in its closure.

The preceding operator range category dichotomy proves continuous functions are meagre in L1:

J (C ([0, 1])) is of the first category in L^{1} (0, 1) .

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